FOR SEMILINEAR WAVE EQUATIONS
AND APPLICATIONS TO YANG-MILLS EQUATIONS
Yung-fu Fang
Abstract. In this work we are concerned with a local existence of cer-tain semilinear wave equations for which the initial data has minimal regularity. Assuming the initial data are in H1+² and H² for any ² > 0, we prove a local result for the problem using a fixed point argument. The main ingredient is an a priori estimate for the quadratic nonlinear term
uDu. They can be applied to the Yang-Mills equations in the Lorentz
gauge.
0. Introduction.
In this paper, we are interested in deriving a new estimate which en-ables us to establish a local existence result for Yang-Mills equations with minimal assumptions on the regularity of the initial data. For this pur-pose, we want to study the following type of system of semilinear wave equations: ( ¤ui = ai jklul∂juk, (x, t) ∈ R3× R, ui(0, x) = fi(x), ui ,t(0, x) = gi(x), x ∈ R3, (0.1)
where ¤ denotes the D’Alembertian −∂2
t + ∆ in R3+1, aijkl are constants,
i, j, k, l = 0, 1, 2, 3, and summation over repeating indices is implied.
For equations (0.1), the energy is E = R |∂tui|2 + |∇ui|2dx, so that the
ideal minimal regularity for initial data is fi ∈ H1(R3), gi ∈ L2(R3).
Unfortunately, this is not the case, see [13] and [14].
In 1979, Segal [17] established a local existence result of classical solu-tion for the Cauchy problem of Yang-Mills equasolu-tions in temporal gauge, with initial data in C3. In 1981, Glassey and Strauss [7] found a class of
global weak solutions of Yang-Mills equations in the temporal gauge with initial data which are radial symmetric. In 1982, Eardley and Moncrief [3] obtained the global existence of a classical solution for the Cauchy problem of Yang-Mills equations in temporal gauge, with initial data in
H2+k(R3) and H1+k(R3) for k ≥ 3. In 1990, Schirmer [16] proved global
existence of spherically symmetric solutions. In 1993, Ponce and Sideris [15] proved a local existence result for a similar equation with the ini-tial data in H2+²(R3) × H1+²(R3) for ² > 0. In 1995, Klainerman and
Machedon [12] proved global existence for the Yang-Mills equations in the temporal gauge with finite energy initial data in R3.
We will prove a local existence result for the system (0.1) assuming that the regularity of initial data is just a little better than energy norm, namely fi ∈ H1+²(R3) and gi ∈ H²(R3), where ² > 0. This result is
optimal ( see [14] Lindblad). The main ingredients in the proof are the standard energy estimate and an a priori estimate given by the following theorem.
Theorem A. (A Priori Estimate) Let ui be a solution of the equations
½
¤ui = bi, (x, t) ∈ R3× R,
ui(0, x) = fi(x), ui
,t(0, x) = gi(x), x ∈ R3.
(0.2)
Then, for every 1
2 > ² > 0, the following estimate holds
³ Z T 0 kD²(ulDuk)kqL2(R3)dt ´1 q 6 C(²) ³ kflkH1+²+ kglkH² + Z T 0 kD²blkL2dt ´ · ³ kfkkH1+² + kgkkH²+ Z T 0 kD²bkkL2dt ´ , (0.3) where i, l, k = 0, 1, 2, 3, q = 2
1−2², the constant C(²) = O(1/
√
²), and D² is a fractional derivative defined via Fourier transform.
Next, we want to apply the above ideas to the Yang-Mills equations which can schematically be written as a system of the following form:
where Aa
µ are the gauge potentials, a = 1, 2, 3, and µ = 0, 1, 2, 3, and
DA represents any kind of first order derivatives of A. For the Yang-Mills
equations in the Lorentz gauge, we can prove a local existence assuming that initial data is in H1+²(R3), for each 1
2 > ² > 0.
1. Semilinear Wave Equation.
Consider the system equations of the following form. ( ¤ui = ai jklul∂juk, (x, t) ∈ R3 × R, ui(0, x) = fi(x), ui ,t(0, x) = gi(x), x ∈ R3. (1.1)
Let us introduce some notations. The space Hs(Rn) is defined to be
Hs(Rn) = {f |(1 − ∆)s2f ∈ L2(Rn)}, (1.2)
where 1 − ∆ is the multiplier operator associated with 1 + |ξ|2. The space
is equipped with the norm k · kHs(Rn) defined by
kf kHs(Rn)= k(1 + |ξ|2) s
2f kˆ L2(Rn).
The space ˙Hs(Rn) is defined to be
˙
Hs(Rn) = {f |(−∆)s2f ∈ Lˆ 2(Rn)}, (1.3)
with norm
kf kH˙s(Rn) = k|ξ|sf kˆ L2(Rn).
The fractional derivative D², via Fourier transform, is defined by
d
D²f (ξ) = |ξ|²f (ξ).ˆ (1.4)
We use the notation D to represent the all possible partial derivatives ∂t
Theorem 1.1. There exists a T > 0 and C depending only on kfik H1+²
and kgik
H², where ² > 0, such that the system of equations (1.1) has
solutions ui : R3 × [0, T ] → R3 with sup [0,T ] kui(t)kH1+²(R3)6 C, (1.5) where i = 0, 1, 2, 3, and Z T 0 kD²(ulDuk)kqL2(R3)dt 6 C, (1.6) where l, k = 0, 1, 2, 3 and q = 2 1 − 2².
To prove theorem 1.1, we need the following main estimate whose proof will be given at the end.
Theorem 1.2. Let ui be the solutions of the system
½ ¤ui = bi, (x, t) ∈ R3× R ui(0, x) = fi(x), ui ,t(0, x) = gi(x), x ∈ R3 (1.7) If fi ∈ H1+²(R3), gi ∈ H²(R3), and 0 < ² < 1 2, then we have ³ Z T 0 kD²(ulDuk)kqL2(R3)dt ´1 q 6 C(²) ³ kflkH1+²+ kglkH² + Z T 0 kD²blkL2dt ´ · ³ kfkkH1+² + kgkkH²+ Z T 0 kD²bkkL2dt ´ , (1.8) where i, l, k = 0, 1, 2, 3, q = 1−2²2 and C(²) = O¡1/√²¢.
We also need the standard energy estimate which states below.
Lemma 1.3. Let ui be the solutions of the system (1.7) and ² > 0. If the
initial data fi∈ H1+²(R3) and gi ∈ H²(R3), then the following estimate
holds sup [0,T ] kui(t)kH1+²(R3) 6 C ³ kfikH1+²+kgikH²+ Z T 0 kD²bikL2(R3)dt ´ . (1.9)
Proof. We omit indices of the ui for simplicity. It suffices to prove the followings: kDu(t, ·)kL2(R3) ≤ C ³ kDu(0, ·)kL2 + Z t 0 kb(τ, ·)kL2dτ ´ , (1.10) ku(t, ·)kL2(R3) ≤ C ³ ku(0, ·)kL2 + Z t 0 kDu(τ, ·)kL2dτ ´ . (1.11) For the estimate (1.10), multiplying (1.7) by ut and integrating by parts
on [0, T ] × R3, then we have Z t=T e(u)dx − Z t=0 e(u)dx = Z T 0 Z R3 b(τ, x)utdxdt, (1.12) where e(u) = 1 2(u2t + |∇u|2).
Let E(t) = Rt=te(u)dx, then we have dE(t) dt = Z R3 butdx ≤ Ckb(t, ·)kL2E 1 2(t). (1.13)
Dividing both sides by E12 and integrating them on [0, t] give (1.10). In
order to prove the estimate (1.11), we use the formula
u(t) − u(0) =
Z t
0
∂tudτ. (1.14)
Taking L2-norm on both sides and applying Minkowski inequality, we get
the desired result. ¤
Now we come to the stage to prove the Theorem 1.1.
Proof of theorem 1.1. The existence proof is based on a simple itera-tion procedure. Let ui
−1 ≡ 0. Define
¤uin = aijkluln−1∂jukn−1, (1.18)
with initial data given by
for all n ≥ 0. Our goal is to prove that u = lim un exists and satisfies the
equation and the desired estimates. It suffices to prove that there exists
T ≤ 1, 4 ≥ 1, such that, for all n,
sup [0,T ] kuin(t) − uin−1(t)kH1+²(R3) ≤ 4 2n+1, (1.20) sup [0,T ] kuin(t)kH1+²(R3) ≤ C, (1.21)
for some constant C that depends on 4 but not on n. Also Z T 0 ° °D²[ul n∂j(ukn− ukn−1)] ° °q L2(R3)dt ≤ 42q 2nq , (1.22) Z T 0 ° °D²[(ul n − uln−1)∂jukn−1] ° °q L2(R3)dt ≤ 42q 2nq , (1.23) Z T 0 ° °D²(ul n∂jukn) ° °q L2(R3)dt ≤ 4 2q. (1.24)
For n = 0, it is easy to check that (1.20)-(1.24) are satisfied.
Next, we assume they are true for some n ≥ 0 and prove (1.20)-(1.24) are true for n + 1 with a T ≤ 1, T is independent of n and the same 4.
C = C(kfik
H1+², kgikH²).
Using (1.24) and (1.8) give ³ Z T 0 ° °D²(ul n+1∂jukn+1) ° °q L2(R3)dt ´1/q ≤C ³ kflkH1+²+ kglkH² + Z T 0 kD²blnkL2dt ´ · ³ kfkkH1+² + kgkkH² + Z T 0 kD²bknkL2dt ´ ≤ 4 2 + CT 1/p ÃZ T 0 ° °D²(ur nDusn) ° °q L2dt !1/q 2 ≤³ 4 2 + CT 1/p4 · 4´2 ≤ 42, where 1 p + 1
Since
bkn− bkn−1 =¡urnDusn− urn−1Dusn¢+¡urn−1Duns − urn−1Dusn−1¢,
thus we can use (1.22) and (1.8) to get ÃZ T 0 ° °D²[ul n+1∂j(ukn+1 − ukn)] ° °q L2(R3)dt !1 q ≤C ³ kflkH1+² + kglkH² + Z T 0 kD²blnkL2dt ´³ Z T 0 ° °D²(bk n− bkn−1) ° ° L2dt ´ ≤C³ 4 2 + T 1/p42´·³ Z T 0 ° °D²£ur n−1D(usn − usn−1) ¤°° L2dt+ Z T 0 ° °D²£(ur n− urn−1)Dusn] ° ° L2dt ´ ≤4hCT1/p(42 2n + 42 2n ) i ≤ 42 2n+1.
The computation for (1.23) is analogous. Using (1.20) and (1.9) give
sup [0,T ] kuin+1(t) − uin(t)kH1+²(R3) ≤C³ Z T 0 kD²(bi n− bin−1)kL2dt ´ ≤ 2CT1/p42 2n ≤ 4 2n+2.
Using (1.21) and (1.9) give
sup [0,T ] kuin+1(t)kH1+²(R3) ≤ C ³ kfikH1+² + kgikH² + Z T 0 kD²binkL2(R3)dt ´ ≤ 4 2 + CT 1/p42 ≤ 4.
Therefore the sequence of functions©ui n
ª
forms a Cauchy sequence under
the norm H1+²(R3). ¤
Proposition 1.4. If f has compact support in the ball BR, then for any
² > 0 there is a constant C such that
kf kL2(R3) 6 C(R, ²)kD²f kL2(R3). (1.25)
Proof . Using Plancherel theorem, we get ° °D²f°° L2 = ° °|ξ|²fˆ°° L2.
and since the Fourier transform of 1
|x|3−² is |ξ|1² so that the function f can
be written as follows. f = F−1³ 1 |ξ|²Dd²f ´ = 1 |x|3−² ∗ D ²f. (1.26)
Hence, from [18], we have
kf k
L3−2²6 (R3) 6 C(²)kD ²f k
L2(R3). (1.27)
Since 3−2²6 > 2 and f has compact support in BR, we have
kf kL2(R3) 6 C(R, ²)kD²f kL2(R3).
¤
Remark. Thus kf kH²(R3) is equivalent to kD²f kL2(R3) if f has compact
support. Therefore, the proof in this paper also works for the ˙Hs(R3)
norm .
2. An Application to Yang-Mills Equations.
In this part, we apply the estimates (1.8) and (1.9) to prove a local existence result for the Yang-Mills equations in the Lorentz gauge.
Let Aµ be the gauge potentials, where µ = 0, 1, 2, 3. The Minkowski
metric is ηµν = diag(−1, 1, 1, 1) and the gauge fields are
where [Aµ, Aν] denotes the commutator. The Lagrangian for the gauge
field Fµν is
L(A) = Z
FµνFµνdxdt.
Here Aµ are regarded as vector valued functions from R4 −→ R3 i.e.
Aµ = (A1µ, A2µ, Aµ3), where Aaµ are functions from R4 to R, and [Aµ, Aν] =
Aµ∧ Aν, ∧ means the usual cross product. For computation need, let us
introduce the notations, ~A = (A1, A2, A3), A = (A0, ~A).
The Yang-Mills equations are the Euler-Lagrange equations of the for-mal Lagrangian
L = 1
2 Z
|E|2 − |B|2dxdt,
subject to the constraints B = −∇ ∧ ~A, E = ∂tA + ∇A~ 0. They can be
written as ¤A0 + ∂0(∂kAk− ∂0A0) + h A0∧ ∂kAk+ ∂kA0 ∧ Ak + Ak ∧ ∂0Ak i + Ak ∧ (A0∧ Ak) = 0, ¤Aj + ∂j(∂kAk− ∂0A0) + h ∂0A0∧ Aj + 2A0 ∧ ∂0Aj − A0 ∧ ∂jA0 + Aj ∧ ∂kAk + (2∂kAj ∧ Ak+ Ak∧ ∂jAk) i + h A0 ∧ (A0∧ Aj) + Ak ∧ (Aj ∧ Ak) i = 0. (2.1) In the Lorentz gauge : −∂0A0 + ∇ · ~A = 0, the Yang-Mills equations
can be rewritten as ¤A0 + A0 ∧ ∂0A0+ 2∂kA0∧ Ak+ Ak∧ ∂0Ak+ Ak∧ (A0 ∧ Ak) = 0, ¤Aj + h 2A0 ∧ ∂0Aj + ∂jA0 ∧ A0 + 2∂kAj ∧ Ak+ Ak∧ ∂jAk i + h A0∧ (A0 ∧ Aj) + Ak∧ (Aj ∧ Ak) i = 0, (2.2) with initial conditions Aµ(0, x) = fµ(x), ∂tAµ(0, x) = gµ(x).
Theorem 2.1. Assume that the initial data fµ ∈ H1+²(R3) and gµ ∈
H²(R3). Then, for any 0 < ² < 1
the initial data such that the system (2.2) has solutions Aµ : R3 × [0, T ] → R3 with sup [0,T ] kAµ(t, ·)kH1+²(R3) ≤ C, (2.4) Z T 0 ° °D²(A µ∧ DAα) ° °q L2(R3)dt 6 C, (2.5) and sup [0,T ] ° °D²[A µ∧ (Aµ∧ Aα)](t, ·) ° ° L2(R3) ≤ C. (2.6)
In order to prove the above theorem, we need the following lemma. Lemma 2.2. (Ponce & Sideris) If f, g ∈ S(Rn) and 1
pi
+ 1
qi
= 1
2, i = 1,
2, with 2 < pi ≤ +∞, then ,for s > 0,
kf gkHs ≤ C ³ kf kLp1 ° °(1 − ∆)s 2g ° ° Lq1 + kgkLp2 ° °(1 − ∆)s 2f ° ° Lq2 ´ . (2.7)
The proof of the theorem 2.1 is similar to that of theorem 1.1. Besides those estimates (1.8), and (1.9), we also need an estimate for the cubic terms in the equations (2.2).
Theorem 2.3. If f, g, and h are all in H1+²(R3), then for any ² > 0
there is a constant C depends on ² such that
kD²(f gh)kL2(R3) ≤ Ckf kH1+²(R3)kgkH1+²(R3)khkH1+²(R3). (2.8)
Proof . Here we employ (2.7), H¨older inequality, Sobolev inequality and (1.25) to estimate D²(f gh) in L2-norm. kD²(f gh)kL2 ≤ kf ghkH² ≤kf kL6 ° °(1 − ∆)²/2(gh)°° L3 + kghkL3 ° °(1 − ∆)²/2f°° L6 ≤kf kH1+² ° °(1 − ∆)²/2(gh)°° H12 + kgkL6khkL6 ° °(1 − ∆)²/2f°° H1 ≤kf kH1+²kghk H12+² + kgkH1+²khkH1+²kf kH1+².
In the same vein, we estimate the product gh in H12+²-norm, kghk H12+² ≤ kgkL6 ° °(1 − ∆)1 4+²2h ° ° L3 + khkL6 ° °(1 − ∆)1 4+²2g ° ° L3 ≤ kgkH1+² ° °(1 − ∆)1 4+2²h ° ° H12 + khkH1+²kgkH1+² ≤ 2kgkH1+²khkH1+². Hence, we have kD²(f gh)kL2 ≤ Ckf kH1+²kgkH1+²khkH1+². ¤ Proof of theorem 2.1. The proof is also based on an iteration scheme like that of theorem 1.1. Define An
µ by setting A−1µ ≡ 0 and for n ≥ 0,
¤Anµ = Fµ(An−1, ∂An−1), (2.9)
with initial conditions
Anµ(o, x) = fµ(x), ∂tAnµ(0, x) = gµ(x). (2.10)
It suffices to prove that there exists T ≤ 1, 4 ≥ 1 such that, for all n, sup [0,T ] ° °(An µ− An−1µ )(t, ·) ° ° H1+² ≤ 4 2n+1, (2.11) sup [0,T ] kAnµ(t, ·)kH1+² ≤ 4, (2.12) Z T 0 ° ° °D²£Anµ ∧ D(Aαn− An−1α )¤°°°q L2(R3)dt ≤ 42q 2nq , (2.13) Z T 0 ° ° °D²£(Anµ− An−1µ ) ∧ DAn−1α ¤°°°q L2(R3)dt ≤ 42q 2nq , (2.14) sup [0,T ] ° ° °D²£Anµ∧¡Aµn∧ (Anα− An−1α )¢¤(t, ·) ° ° ° L2(R3) ≤ 43 2n , (2.15) sup [0,T ] ° ° °D²£Anµ∧¡(Anµ− An−1µ ) ∧ An−1α ¢¤(t, ·) ° ° ° L2(R3) ≤ 43 2n , (2.16) sup [0,T ] ° ° °D²£(Anµ− An−1µ ) ∧ (An−1µ ∧ An−1α )¤(t, ·) ° ° ° L2(R3) ≤ 43 2n , (2.17)
ÃZ T 0 ° ° °D²(An µ∧ DAnα) ° ° °q L2(R3)dt !1/q ≤ 42, (2.18) sup [0,T ] ° ° °D²£Anµ∧ (Anµ∧ Anα)¤(t, ·) ° ° ° L2(R3) ≤ 4 3. (2.19)
For n = 0, the above inequalities follow from the energy estimate (1.9) and theorem 1.2. Next, we assume that inequalities (2.11)-(2.19) are true for some n ≥ 0 and prove they are true for n + 1. The calculation here is similar to those in the proof of theorem 1.1. Inequalities (2.13), (2.14) and (2.18) follow from the proof of theorem 1.2. Using theorem 2.3 gives inequalities (2.15)-(2.17), and (2.19) for all n ≥ 0. Using (1.9), (1.8), and (2.8) gives inequalities (2.11), (2.12) for all n ≥ 0. This completes the
proof. ¤
3. Proof of the Estimates.
In this part, we prove theorems 1.2.
Proof of Theorem 1.2. By Duhamel’s principle, it suffices to prove the following case: ¤ui = 0, ui(0, x) = 0, ui
,t(0, x) = fi(x). For the case, D
= ∂j, since ¯ ¯ ¯ηj |η| ¯ ¯
¯ ≤ 1, it is enough to prove the estimate for D = ∂t. Let
A(ξ) = 2πi(t|ξ| + x · ξ), B(ξ) = 2πi(−t|ξ| + x · ξ).
For simplicity, we skip the superscripts. Then the solution and its time derivative can be written as
u(t, x) = Z 1 |ξ| ¡ eA(ξ)− eB(ξ)¢f (ξ)dξ = uˆ +(t, x) − u−(t, x), (3.3a) ∂tu(t, x) = Z ¡ eA(η)+ eB(η)¢f (η)dη = ∂ˆ tu+(t, x) + ∂tu−(t, x).(3.3b)
Then for u+∂tu+, we have
u+∂tu+ =
Z Z 1
|ξ|e
Consider its inner product with an arbitrary function g, D D²(u+∂tu+), g E = Z Z |ξ + η|² |ξ| ˆgef (ξ) ˆˆ f (η)dξdη, (3.5)
where ˆge = ˆg(−|ξ| − |η|, −ξ − η). Taking absolute value on both sides, we
can get the inequality ¯ ¯ ¯ D D²(u+∂tu+), g E¯ ¯ ¯ ≤ ³ Z Z |ξ + η|2² |ξ|2+2²|η|2²|ˆge| 2dξdη´ 1 2 kf k2H². (3.6)
Let ξ = ρω where ω is a unit vector, ξ + η = z, and |ξ| + |η| = τ. Then we have the formula for changes of variables
ρ = 1 2 τ2− |z|2 τ − z · ω, (3.7a) |z − ρω| = 1 2 τ2− 2τ z · ω + |z|2 τ − z · ω , (3.7b) dρ dτ = 1 2 τ2− 2τ z · ω + |z|2 (τ − z · ω)2 . (3.7c) Since dξdη = dξdz = ρ2dρdωdz = ρ2dρ dτdωdτ dz,
we can change the variables in the integral (3.6) to get ¯ ¯ ¯ D D²(u+∂tu+), g E¯¯ ¯ ≤ ³ Z Z Z |z|2² ρ2+2²|z − ρω|2²ρ 2dρ dτdω|ˆg(−τ, −z)| 2dτ dz´ 1 2 kf k2H². We denote E(τ, z) = Z |z|2² ρ2+2²|z − ρω|2²ρ 2dρ dτdω. (3.8)
Employ (3.7) we can estimate it as follows.
E(τ, z) = Z |z|2² ρ2²|z − ρω|2² dρ dτdω ≤ τ 2² Z 1 ρ2²|z − ρω|2² dρ dτdω = τ 2² (τ2 − |z|2)2² Z (τ2+ |z|2− 2τ z · ω)1−2² (τ − z · ω)2−4² dω ≤ C τ2² 1 (1 − λ)2² Z π 0 (1 + λ2 − 2λ cos θ)1−2² (1 − λ cos θ)2−4² sin θdθ ≤ C τ2² Z 1+λ 1−λ 1 (s − 1+λ2 )1−2² s2−4² ds ≤ C(²) τ2²(1 − λ)2² ≤ C(²) (τ − |z|)2²,
where the used substitution λ = |z| τ , z · ω = |z| cos θ and s = 1 − λ cos θ 1 − λ . Here ² > 0 and C(²) = O(1 ²). Therefore, we have ¯ ¯ ¯ D D²(u+∂tu+), g E¯¯ ¯ ≤ C(²) µZ Z 1 |τ − |z||2²|ˆg(−τ, −z)| 2dτ dz ¶1 2 kf k2H², (3.9) where C(²) = O(√1
²). Now we want to estimate the integral
Z Z
1
|τ − |z||2²|ˆg(−τ, −z)|
2dτ dz.
Since we know that if 0 < 2² < 1, the Fourier transform of 1
|τ |2² is
C²
|t|1−2²,
where C² is some constant depends on ² (see [18] Stein). Using
Hardy-Littlewood inequality, we have the following Z Z 1 |τ − |z||2²|ˆg(−τ, −z)| 2dτ dz =D 1 |τ − |z||2²g, ˆˆ g E =DFt−1³ 1 |τ − |z||2² ´ ∗ Fx(g), Fx(g) E = C²D e it|z| |t|1−2² ∗ Fx(g), Fx(g) E =C² Z Z Z ei(t−s)|z| |t − s|1−2²Fx(g)(s, −z)Fx(g)(t, −z)dsdtdz ≤C² Z Z 1 |t − s|1−2²kFx(g)(s, ·)kL2kFx(g)(t, ·)kL2dsdt ≤C² µZ kg(t, ·)kpL2dt ¶2 p , where p = 2 1 + 2², F −1
t is the inverse Fourier transform over time t and
Fx the Fourier transform over space variable x. Therefore, we have
¯ ¯ ¯ D D²(u+∂tu+), g E¯ ¯ ¯ ≤ C µZ kg(t, ·)kpL2dt ¶1 p kf k2H². (3.10) Hence, we get µZ ° ° °D²(ul+Duk+) ° ° °q L2(R3)dt ¶1/q ≤ C(²)kflkH²kfkkH², (3.11)
where C(²) = O(√1
²) and q =
2
1 − 2². The calculation for the term
u−∂tu− is analogous to that of the above.
For the term, u+∂tu−, we have
D D²(u+∂tu−), g E = Z Z Z Z |ξ + η|² |ξ| e A(ξ)+B(η)f (ξ) ˆˆ f (η)dξdηg(t, x)dtdx = Z Z |ξ + η|2² |ξ| ˆghf (ξ) ˆˆ f (η)dξdη (3.12)
We can split the above integral into two parts, one is |ξ| > |η| and another is |ξ| < |η|. Let D D²(u+∂tu−), g E + = Z Z {|ξ|>|η|} |ξ + η|² |ξ| ˆghf (ξ) ˆˆ f (η)dξdη, (3.13) and D D²(u+∂tu−), g E − = Z Z {|ξ|<|η|} |ξ + η|² |ξ| ˆghf (ξ) ˆˆ f (η)dξdη. (3.14)
For (4.13), using the change of variables, we can rewrite it as D D²(u+∂tu−), g E + = Z Z Z Σ |z|² ρ f (ρω) ˆˆ f (z − ρω)ρ 2dρ dτdωˆg(−τ, −z)dτ dz, (3.15) where Σ = {ω : |ω| = 1, |z| ≥ ω · z ≥ τ > 0}, ˆgh = ˆg(−|ξ| + |η|, −ξ − η),
ξ = ρω, z = ξ + η, and τ = |ξ| − |η|. It is sufficient to consider τ > 0.
Thus, we have the formula for changes of variables that
ρ = 1 2 |z|2 − τ2 z · ω − τ, (3.16a) |z − ρω| = 1 2 |z|2 − 2τ z · ω + τ2 z · ω − τ , (3.16b) dρ dτ = 1 2 |z|2 − 2τ z · ω + τ2 (z · ω − τ )2 , (3.16c)
and the Fourier transform of ∂²(u +∂tu−) is F(∂²(u+∂tu−))(−τ, −z) = Z Σ |z|² |ρ| f (ρω) ˆˆ f (z − ρω)ρ 2dρ dτdω. (3.17) Let Σ = Σ1 ∪ Σ2, which Σ1 = {z : |z| ≥ ω · z ≥ τ + δ(1 − λ)|z|}, and Σ2 = {z : τ + δ(1 − λ)|z| ≥ ω · z ≥ τ > 0},
where δ is any small positive number. We will estimate the integral on Σ1 and Σ2 separately. Thus we can split the integral into two parts so
that D D²(u +∂tu−), g E + = Z Z Z Σ1 · · · + Z Z Z Σ2 · · · = I1 + I2. (3.18)
For the term I1, we have
I1 = Z Z Z Σ1 |z|² |ρ|f (ρω) ˆˆ f (z − ρω)ρ 2dρ dτdωˆg(−τ, −z)dτ dz. Then |I1| ≤ µZ Z Z Σ1 |z|2² ρ2+2²|z − ρω|2²ρ 2dρ dτdω|ˆg(−τ, −z)| 2dτ dz ¶1 2 kf k2H². (3.19) We denote the notation
HΣ1(τ, z) = Z Σ1 |z|2² ρ2+2²|z − ρω|2²ρ 2dρ dτdω.
Thus, we can use (3.16) and some substitution to get
HΣ1(τ, z) = Z Σ1 |z|2² ρ2+2²|z − ρω|2²ρ 2dρ dτdω = |z| 2² Z 1 ρ2²|z − ρω|2² dρ dτdω = |z| 2² (|z|2− τ2)2² Z (τ2 + |z|2 − 2τ z · ω)1−2² (z · ω − τ )2−4² dω ≤ C |z|2²(1 − λ)2² Z cos−1(λ+δ(1−λ)) 0 (1 + λ2− 2λ cos θ)1−2² (cos θ − λ)2−4² sin θdθ ≤ C |z|2² Z 1 δ (1 + λ − 2λs)1−2² s2−4² ds < C(δ) |z|2² < C(δ) (|z| − τ )2²,
where we used the substitution λ = τ
|z|, z·ω = |z| cos θ, and s =
cos θ − λ 1 − λ . Here C(δ) = C
(1 − 4²)δ1−4² and C is some constant. Therefore, we get
|I1| ≤ µZ Z C(δ) ||z| − τ |2²|ˆg(−τ, −z)| 2dτ dz ¶1 2 kf k2H². (3.20)
Now we want to estimate I2. In order to do this we need two things.
First, we note that on Σ2 there exists constants C1, C2, such that
C1 < |ξ| |η| < C2. (3.21) This is because |ξ| |η| = ρ |z − ρω = |z|2 − τ2 |z|2+ τ2 − 2τ z · ω = 1 − λ2 1 + λ2− 2λ cos θ, and λ < cos θ < λ + δ(1 − λ) on Σ2. Hence, if we choose δ = 1 2, then we get 1 < |ξ| |η| < 2. We also have |ξ + η| ≤ |ξ| + |η| ≤ C|ξ| (or C|η|). Second, let’s consider the following
system of equations ½
¤φ = 0,
φ(0) = 0, ∂tφ(0) = f0,
(3.22) where f0 satisfies ˆf0 = | ˆf |, then we have the result
° ° °(D12+²φ±)2 ° ° ° L2(R4) ≤ C ° ° °D²f0 ° ° °2 L2(R3), (3.23)
for any ² ≥ 0. The proof of (3.23) can be done directly in the same manner as above and will be given outline later.
For the term I2, employing (3.21) and (3.23), we have
|I2| ≤ C Z Z 1 |z|² Z Σ2 |z|²|z|² |ρ| ¯ ¯ ¯ ˆf (ρω) ¯ ¯ ¯ ¯ ¯ ¯ ˆf (z − ρω) ¯ ¯ ¯ρ2dρ dτdω ¯ ¯ ¯ˆg(−τ, −z) ¯ ¯ ¯dτ dz ≤ C Z Z 1 |z|² Z Σ2 1 |ρ|12−²|z − ρω|12−² ¯ ¯ ¯ ˆf (ρω) ¯ ¯ ¯ ¯ ¯ ¯ ˆf (z − ρω) ¯ ¯ ¯ρ2dρ dτdω|ˆg|dτ dz
The integral over Σ2is basically the Fourier transform of D 1
2+²φ+D12+²φ−
which can be seen as follows. D D12+²φ+D12+²φ−, g E = Z Z 1 |ξ|12−²|η|12−² ˆ g(−|ξ| + |η|, −ξ − η) ˆf0(ξ) ˆf0(η)dξdη = Z Z Z Σ 1 |ρ|12−²|z − ρω|12−² ˆ f0(ρω) ˆf0(z − ρω)ρ2dρ dτˆg(−τ, −z)dτ dz.(3.24)
Therefore we can estimate I2 as follows.
|I2| ≤ C Z Z 1 |z|²F ³ D12+²φ+D12+²φ− ´ |ˆg|dτ dz ≤ C µZ Z 1 |z|2²χ{0<τ <|z|}|ˆg| 2dτ dz ¶1 2 °° °D12+²φ+D12+²φ− ° ° ° L2(R4) ≤ C µZ Z 1 ||z| − τ |2²|ˆg| 2dτ dz ¶1 2 °° °D²f0 ° ° °2 L2.
Combining the estimates for the terms I1 and I2, we have
¯ ¯ ¯ D D²(u+∂tu−), g E¯ ¯ ¯ ≤ C(², δ) µZ Z 1 ||z| − τ |2²|ˆg| 2dτ dz ¶1 2 kf k2H². (3.25)
Thus, with the same reason as in the estimate of the previous term,
D²(u +∂tu+), we have ¯ ¯ ¯ D D²(u+∂tu−), g E¯ ¯ ¯ ≤ C µZ kg(t, ·)kpL2dt ¶1 p kf k2H². Hence, we get µZ ° ° °D²(ul+Duk−) ° ° °q L2(R3)dt ¶1/q ≤ C(²)kflkH²kfkkH², (3.26) where q = 2 1 − 2².
Finally, we will discuss the estimate of hD²(u
+∂tu−), gi− briefly. D D²(u+∂tu−), g E − = Z Z {|ξ|<|η|} |ξ + η|² |ξ| ˆghf (ξ) ˆˆ f (η)dξdη. (3.27)
Let’s switch the variables ξ with η, thus we get D D²(u +∂tu−), g E − = Z Z Z Σ |z|² |z − ρω|f (ρω) ˆˆ f (z − ρω)ρ 2dρ dτdωˆg(−τ, −z)dτ dz.
For this case, we compute HΣ1 as follows.
HΣ1(τ, z) = Z Σ1 |z|2²ρ2−2² |z − ρω|2+2² dρ dτdω ≤ C |z|2² Z 1 δ 1 (1 + λ − 2λs)1+2²s2−4²ds ≤ C |z|2² µ 1 2² 1 (1 − λ)2² + 1 1 − 4² 1 δ1−4² ¶ ≤ C(², δ) (|z| − τ )2².
Thus, it works for the term, I1. For another term, I2, the proof is
sim-ilar. The proof for the term D²(u
−∂tu+) is similar to those of the term
D²(u
+∂tu−). Hence, the proof is complete. ¤
Proof of (3.23). We only outline the proof. Since ³ D12+²φ+ ´2 = Z Z 1 |ξ|12−²+ |η|12−² eA(ξ)+A(η)fˆ0(ξ) ˆf0(η)dξdη, thus we have ¯ ¯ ¯D¡D12+²φ+¢2, g E¯¯ ¯2 ≤ Z Z Z 1 ρ|z − ρω|ρ 2dρ dτdω|ˆg(−τ, −z)| 2dτ dz°°°D²f0°°°4 L2(R3).
The inner-most integral called E(τ, z) can be computed as follows.
E(τ, z) = Z 1 ρ|z − ρω|ρ 2dρ dτdω = Z π 0 1 − λ2 (1 − λ cos θ)2 sin θdθ = 1 + λ λ Z 1+λ 1−λ 1 1 s2ds
which is bounded by some constant. Hence we have ° ° °(D12+²φ+)2 ° ° ° L2(R4) ≤ C ° ° °D²f0 ° ° °2 L2(R3).
Similar calculation gives ° ° °(D12+²φ−)2 ° ° ° L2(R4) ≤ C ° ° °D²f0 ° ° °2 L2(R3). ¤
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Department of Mathematics, Cheng Kung University, Tainan 701 Taiwan
