AR(ζ0, D)S0(λ)( ˜ζ0F0) 0
0 0
+
∞
X
`=1
AR(ζ`, D)S`(λ)( ˜ζ`F0,ζ˜`G,ζ˜`G0,ζ˜`g100) BR(ζ`, D)S`(λ)( ˜ζ`F0,ζ˜`G,ζ˜`G0,ζ˜`g100) λ1/2BR(ζ`, D)S`(λ)( ˜ζ`F0,ζ˜`G,ζ˜`G0,ζ˜`g001) λBR1(ζ`, D)(S`1(λ)( ˜ζ`F0,ζ˜`G,ζ˜`G0,ζ˜`g100)
.
By (6-1), (6-7) and Proposition 2.2, we have
RL(Xq(Ω))({(τ ∂τ)sU(λ)|λ∈Σϑ,λ2})≤Cqκ1λ−1/22 (s= 0,1) (6-14) for anyλ2≥λ1. We chooseλ2so large that
Cqκ1λ−1/22 ≤1/2. (6-15)
By (6-13), (6-14) and (6-15), we have
kHλU(λ)(F, G)kXq(Ω)≤12kHλ(F, G)kXq(Ω). (6-16) By the equivalence of the norms kHλ(·)kXq(Ω) and k · kGq(Ω) (cf. (5-19)), the inverse (I−U(λ))−1 = I+P∞
n=1U(λ)n exists in L(Gq(Ω)). LetV =A(λ)Hλ(I−U(λ))−1(F, G), and then by (6-12)V solves the equation:
Lλ(D)V = (F, G) on Ω×Γ. (6-17)
The uniqueness of solutions follows from the existence of solutions of the dual problem, so thatV is the unique solution of the equation (6-17).
On the other hand, by (6-14) and (6-15), I− U(λ) =I+P∞
n=1U(λ)n exists in C(Σϑ,λ2,L(Xq(Ω))) and satisfies the estimate:
RL(Xq(Ω))({(τ ∂τ)s(I− U(λ))−1|λ∈Σϑ,λ2})≤2 (s= 0,1). (6-18) Moreover, by (6-13),
Hλ(I−U(λ))−1=Hλ+
∞
X
n=1
HλU(λ)n= (I+
∞
X
n=1
U(λ)n)Hλ= (I− U(λ))−1Hλ. (6-19) LetBi(λ) =Ai(λ)(I− U(λ))−1 (i= 1,2). Then, by (6-9), (6-18) and Proposition 2.2,
RL(X
q(Ω),Hq4−j(Ω))({(τ ∂τ)s(λj/2B1(λ))|λ∈Σϑ,λ1})≤Cqκ1 (s= 0,1, j = 0,1,2,3,4), RL(X
q(Ω),Hq2−j(Ω))({(τ ∂τ)s(λj/2B2(λ))|λ∈Σϑ,λ1})≤Cqκ1 (s= 0,1, j = 0,1,2).
Moreover, settingB(λ) = (B1(λ), λB1(λ),B2(λ))>, by (6-19) we see that B(λ)Hλ(F, G) =A(λ)Hλ(I−U(λ))−1(F, G),
and thereforeV =B(λ)Hλ(F, G) is the unique solution of the equation (1-5) (cf. (6-17)). This completes the proof of Theorem 1.4 replacingλ0byλ2.
7 P roof of Theorem 1.3
To prove Theorem 1.3, we start with
Lemma 7.1. Let 1< p, q <∞. For any θ0∈Bq,p2−2/p(Ω),u0 ∈Bq,p4−2/p(Ω), and u1 ∈B2−2/pq,p (Ω), there exist ω andwsuch that ω|t=0=θ0,w|t=0=u0, and∂tw|t=0=u1 inΩ,
ω∈
1
\
`=0
Hp`((0,∞), Hq,p2−2`(Ω)), w∈
2
\
`=0
Hp`((0,∞), Hq4−2`(Ω),
and
1
X
`=0
k∂t`ωkL
p((0,∞),Hq2−2`(Ω)≤Ckθ0kB2−2/p q,p (Ω),
2
X
`=0
k∂t`wkL
p((0,∞),Hq,p4−2`(Ω)≤C(ku0kB4−2/p
q (Ω)+ku1kB2−2/p q,p (Ω)).
with some positive constantC.
Proof. To prove the lemma, we consider the shifted heat equation:
∂tv+v−∆v=f in (0,∞)×RN, v|t=0=v0. (7-1) Employing the same argument as in the previous sections, it is easy to prove that for anyf ∈Lp((0,∞), Lq(RN)) andv0∈B2−2/pq,p (RN), problem (7-1) admits a unique solutionv∈T1
`=0Hp`((0,∞), Hq2−2`(RN)) possessing the estimate:
1
X
`=0
k∂t`vkL
p((0,∞),Hq2−2`(RN))≤C(kv0kB2−2/p
q,p (RN)+kfkLp((0,∞),Lq(Ω))). (7-2) Assume thatv0∈B4−2/pq,p (RN) andf ∈T1
`=0Hp`((0,∞), Hq2−2`(RN)). Then, for any multi-indexα∈NN0
with|α| ≤2 we have
∂t∂xαv+∂xαv−∆∂xαv=∂xαf in (0,∞)×RN, ∂xαv|t=0=∂xαv0. Thus, the unique existence of solutions of (7-1) yields that
v∈
1
\
`=0
Hp`((0,∞), Hq4−2`(RN)),
1
X
`=0
k∂t`vkL
p((0,∞),H4−2`q (RN))≤C(kv0kB4−2/p
q,p (RN)+kfkLp((0,∞),H2 q(RN))).
(7-3)
Moreover, the relation: ∂t2v=−∂tv+ ∆∂tv+∂tf yields that∂t2v∈Lp((0,∞), Lq(RN)) and k∂2tvkLp((0,∞),Lq(RN))≤C(kv0kB4−2/p
q,p (RN)+
1
X
`=0
k∂t`fkL
p((0,∞),Hq2−2`(RN))). (7-4) Letιhbe an extension map as given in Introduction satisfying property (e-1). Letω be a solution of the shifted heat equation:
∂tω+ω−∆ω= 0 in (0,∞)×RN, ω|t=0=ιhθ0. Since ιhθ0 ∈Bq,p2−2/p(RN), by (7-2) with f = 0 there exists a uniqueω ∈T1
`=0Hp`((0,∞), Hq2−2`(RN)) satisfying the estimate
1
X
`=0
k∂`tωkL
p((0,∞),Hq2−2`(RN))≤Ckιhθ0kB2−2/p q,p (RN). Sinceιhθ0=θ0 on Ω andkιhθ0kB2−2/p
q,p (RN)≤Ckθ0kB2−2/p
q,p (Ω) with some constantC >0, the restriction ofω on Ω is the function satisfying the required properties.
Next, let f be a solution of the shifted heat equation:
∂tf+f−∆f = 0 in (0,∞)×RN, f|t=0=v0
withv0=ιhu1+ιhu0−ιh∆u0. Sincev0∈Bq,p2−2/p(RN) and kv0kB2−2/p
q,p (RN)≤C(ku1kB2−2/p
q,p (Ω)+ku0kB4−2/p q,p (Ω)), there exists a uniquef ∈T1
`=0Hp`((0,∞), Hq2−2`(RN)) satisfying the estimate:
1
X
`=0
k∂t`fkL
p((0,∞),Hq2−2`(RN)) ≤C(ku1kB2−2/p
q,p (Ω)+ku0kB4−2/p
q,p (Ω)). (7-5)
Letwbe a solution of the shifted heat equation:
∂tw+w−∆w=f in (0,∞)×RN, w|t=0=ιu0. (7-6) Then, by (7-3), (7-4) and (7-5), there exists a unique w ∈ T2
`=0Hp`((0,∞), Hq4−2`(RN)) satisfying the estimate:
2
X
`=0
k∂t`wkL
p((0,∞),Hq4−2`(RN)) ≤C(ku1kB2−2/p
q,p (Ω)+ku0kB4−2/p q,p (Ω)).
Moreover, by (7-6),∂tw|t=0 =v0−ιhu0+ ∆ιhu0=ιhu1, so that the restriciton ofwon Ω satisfies the required properties, which completes the proof of Lemma 7.1.
In view of Lemma 7.1, to prove Theorem 1.3, it suffices to consider the equations (1-4) withU0= 0.
LetF = (0, f1, f2)> andG= (g1, g2, g3) satisfy the regularity condition:
(f1, f2)∈Lp((0, T), Lq(Ω)2), G∈Hp1((0, T), Lq(Ω)×W−1q (Ω)2)∩Lp((0, T), Hq2(Ω)×Hq1(Ω)2) and the compatibility condition: G|t=0 = 0. In the following, given f(t,·) defined fort ∈[0, T], f0(t,·) denotes the zero extension off tot <0, that is,f0(t,·) =f(t,·) for 0< t < T, andf0(t,·) = 0 fort <0, andETf denotes the extention of f toRdefined by
[ETf](t,·) =
(f0(t,·) fort < T ,
f0(2T−t,·) fort≥T . (7-7)
Note thatETf vanishes fort6∈[0,2T] and moreover, iff|t=0= 0, then
∂t[ETf](t,·) =
∂tf(t,·) fort≤T ,
−(∂tf)(2T−t,·) fort≥T ,
0 fort6∈[0,2T].
(7-8)
SinceETf =f for 0≤t≤T, instead of the equations (1-4) withU0= 0, we consider the equations:
Ut−A(D)U =ETF inR×Ω, B(D)U =ETG onR×Γ. (7-9) Let L be the Laplace transform with respect to time variable t and let L−1 be its inverse transform, which are defined by
L[f](λ,·) = Z ∞
−∞
e−λtf(t,·)dt= Z
R
e−iτ te−γtf(t,·)dt, L−1[f](t,·) = 1
2πi
Z γ+i∞
γ−i∞
eλtf(λ,·)dλ=eγt 2π
Z ∞
−∞
f(γ+iτ,·)dτ,
whereλ=γ+iτ ∈C. Let L[ETF](λ,·) =Jλ andL[ETG](λ,·) =Kλ. SinceETF and ETGvanish for t 6∈[0,2T], Jλ andKλ are entire functions with respect to λ∈C. Applying the Laplace transform to (7-9), we have
λVλ−A(D)Vλ=Jλ in Ω, B(D)Vλ=Kλ on Γ, (7-10)
where we have setVλ=L[U](λ,·). Now,Vλ=B(λ)Hλ(Jλ, Kλ) is the unique solution of (7-10), so that the uniqueness of the solution yields that Vλ is holomorphic forλ∈Σϑ,λ0. LetBi(λ) (i= 1,2) be the operators given in Theorem 1.4 and setB(λ) = (B1(λ), λB1(λ),B2(λ))>. Then, the unique solutionU of (7-9) is given by
U(t,·) =L−1[B(λ)Hλ(Jλ, Kλ)] =eγtF−1[B(λ)Hλ(F[e−γ·ETF],F[e−γ·ETG])](t,·).
Let Λ1/2 be the operator defined by
[Λ1/2f](t,·) =L−1[λ1/2L[f](λ,·)](t), and let
ke−γtfkLp(R,X)=nZ ∞
−∞
e−γtkf(t,·)kX)pdto1/p .
Since|λ|1/2≤ |λ|(1 +|ξ|2)−1/2 when|λ| ≥1 +|ξ|2 and|λ|1/2≤(1 +|ξ|2)1/2when|λ| ≤1 +|ξ|2, by the Bourgain theorem (cf. Proposition 2.2) we have
keγtΛ1/2fkLp(R,Lq(RN))≤C(keγt∂t(1−∆)−1/2fkLp(R,Lq(RN))+keγt(1−∆)1/2fkLp(R,Lq(RN))), so that by using property (e-2) of the extension mapιhgiven in the introduction we have
keγtΛ1/2fkLp(R,Lq(Ω))≤C(keγt∂tfkL
p(R,W−1q (Ω))+keγtfkLp(R,H1
q(Ω))). (7-11) And also, using the extension mapιh and Proposition 2.2, we have
ke−γt∂tfkLp(R,Hm
q(Ω))+ke−γtfkL
p(R,Hqm+2(Ω))
≤Cm,q(ke−γt∂tfkLp(R,Hm
q (Ω))+ke−γtΛ1/2∇fkLp(R,Hm
q(Ω))+ke−γt∇2fkLp(R,Hm
q (Ω))) (7-12) for anym∈N∪ {0}. Therefore, using (7-11) and (7-12) and applying the Weis operator valued Fourier multiplier theorem with the help of Theorem 1.4, we have
2
X
`=0
ke−γt∂t`ukL
p(R,Hq4−2`(Ω))+
1
X
`=0
ke−γt∂t`θkL
p(R,Hq2−2`(Ω))
≤C(ke−γtET(f1, f2)kLp(R,Lq(Ω))+ke−γtETGkLp(R,H2q(Ω)×Hq1(Ω)2)
+ke−γtΛ1/2ETGkLp(R,Hq1(Ω)×Lq(Ω)2)+ke−γt∂tETg1kLp(R,Lq(Ω)))
≤C(ke−γtET(f1, f2)kLp(R,Lq(Ω))+ke−γtETGkLp(R,H2q(Ω)×Hq1(Ω)2)
+ke−γt∂tETGkL
p(R,Lq(Ω)×W−1q (Ω)2))
(7-13)
for anyγ ≥λ0 with constant C independent of γ, where we have setU = (u, ∂tu, θ). Using (7-7) and (7-8) and noting thatG|t=0= 0, we have
ke−γtET(f1, f2)kLp(R,Lq(Ω))≤Ck(f1, f2)kLp((0,T),Lq(Ω)), ke−γtETGkLp(R,H2
q(Ω)×Hq1(Ω)2)≤CkGkLp((0,T),H2
q(Ω)×Hq1(Ω)2), ke−γt∂tETGkL
p(R,Lq(Ω)×W−1q (Ω)2)≤Ck∂tGkL
p((0,T),Lq(Ω)×W−1q (Ω)2), which, combined with (7-13), furnishes that
2
X
`=0
ke−γt∂t`ukL
p(R,Hq4−2`(Ω))+
1
X
`=0
ke−γtθkL
p(R,Hq2−2`(Ω))≤CIT (7-14) with
IT =ke−γt(f1, f2)kLp((0,T),Lq(Ω))+kGkLp(R,H2q(Ω)×Hq1(Ω)2)+k∂tGkL
p(R,Lq(Ω)×W−1q (Ω)2).
Especially, for anyt <0, we have eγ|t|
2
X
`=0
k∂t`ukL
p((−∞,t),Hq4−2`(Ω))+
1
X
`=0
kθkL
p((−∞,t),Hq2−2`(Ω))
≤
2
X
`=0
ke−γt∂t`ukL
p(R,Hq4−2`(Ω))+
1
X
`=0
ke−γtθkL
p(R,Hq2−2`(Ω))≤CIT
for anyγ≥λ0, so that lettingγ→ ∞, we have eγ|t|
2
X
`=0
k∂t`ukL
p((−∞,t),Hq4−2`(Ω))+
1
X
`=0
kθkL
p((−∞,t),Hq2−2`(Ω))= 0 for anyt <0, which implies that (u, θ) = 0 fort <0.
Summing up, we have proved thatU = (u, ∂tu, θ)> satisifes the equations:
Ut−A(D)U =F in Ω×(0, T), B(D)U =G on Γ×(0, T), U|t=0= 0 in Ω and the estimate:
2
X
`=0
ke−γt∂t`ukL
p((0,T),Hq4−2`(Ω))+
1
X
`=0
ke−γtθkL
p((0,T),H2−2`q (Ω))≤CIT.
The uniqueness of the solutions follwos from the existence of solutions of the dual problem, which completes the proof of Theorem 1.3.
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