SLIDE 1
Thermoelastic plate equation ∂2
t u + ∆2u + ∆θ = f1
in R+ × Ω,
∂tθ − ∆θ − ∆∂tu = f2
in R+ × Ω, u = ∂νu = θ = 0
- n R+ × ∂Ω,
On the L p L q maximal regularity for linear thermoelastic plate - - PowerPoint PPT Presentation
On the L p L q maximal regularity for linear thermoelastic plate - - PowerPoint PPT Presentation
On the L p L q maximal regularity for linear thermoelastic plate equation in a bounded domain Yuka Naito (Sato) Waseda University 25/08/2009 Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 1 / 38
SLIDE 2
SLIDE 3
The reason of the appearance of ∆2 Elasticity
displacement : u = X(t, x) − x X(t, x): position vector.
The case of the plate
Considering the displacement of the center of the plate first, I measure the distance from the center of the plate.
Necessary assumptions
1
The center of the plate is always center.
2
The vertical segment of the center of the plate keeps vertical.
3
The thickness of x3 is so small.
4
The curve of the plate is so small.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 3 / 38
SLIDE 4
The displacement of the plate in the 3 dimension case.
The displacement of the center of the plate: u1(x1, x2), u2(x1, x2), u3(x1, x2). The displacement of the plate U1(x1, x2, x3) = u1(x1, x2) − x3
∂u3 ∂x1 ,
U2(x1, x2, x3) = u2(x1, x2) − x3
∂u3 ∂x2 ,
U3(x1, x2, x3) = u3(x1, x2) + x3.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 4 / 38
SLIDE 5
The type of equations
I shall consider type of equations.
Thermoelastic plate equation ∂2
t u + ∆2u + ∆θ = f1
Dispersive type
∂tθ − ∆θ − ∆∂tu = f2
Parabolic type Thermoelastic plate equation is the couple of these two type. I consider type of the whole equation.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 5 / 38
SLIDE 6
The classification of equations by the characteristic root in a Cauchy problem Parabolic type
Heat eqauton: ∂tτ − ∆τ = 0. −
→ −|ξ|2. Dispersive type
Plate equation: ∂2
t u + ∆2u = 0. −
→ ±i|ξ|2. The couple of these two type
Thermoelastic plate equation
{ ∂2
t u + ∆2u + ∆θ = 0,
∂tθ − ∆θ − ∆∂tu = 0. − → −α|ξ|2, −β|ξ|2, −¯ β|ξ|2 (α : A positive real number, Reβ > 0.)
Therefore I know thermoelastic plate equation is parabolic type. But when I consider a initial boundary problem, the classification of equations is not so obvious.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 6 / 38
SLIDE 7
What is the parabolicity
I shall consider a general initial value problem.
A initial value problem { ∂tu(t) = Au(t)
u|t=0 = u0 A : a general operator, u0 : a initial value.
The definition of parabolic type
u0 ∈ X =
⇒ u(t) ∈ D(Am) (∀m ≥ 1), t > 0
D(A) : a domain of A.
⇐ = Resolvent estimate ∃λ0 ∈ R Reλ ≥ λ0 = ⇒ |λ| ∥(λ − A)−1∥ ≤ C.
In general if it holds resolvent estimate, then it is parabolic.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 7 / 38
SLIDE 8
The definition of maximal regularity
X : Banach space, E : A dense subspace of X, A : E → X ; A closed linear operator.
A Cauchy problem. ∂tu = Au + f (0 < t < T),
u|t=0 = a
(∗)
If (∗) admits a solution u ∈ Lp((0, T), E) ∩ W1
p((0, T), X) = Mp((0, T) : A),
then a ∈ [X, E]1−1/p,p : the trace class of Mp((0, T) : A) of t = 0, f ∈ Lp((0, T), X). A has the Lp-maximal regularity ⇐
⇒ if a ∈ [X, E]1−1/p,p and
f ∈ Lp((0, T), X), then (∗) admits a unique solution u ∈ Mp((0, T) : A) A has the Lp-maximal regularity. =
⇒ A is parabolic type.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 8 / 38
SLIDE 9
The known results about thermoelastic plate equation By Kim, Rivera-Racke, Liu-Zeng, Avalos-Lasiecka, Lasiecka-Triggiani, Shibata
A bounded domain · · · The exponatial decay with the several boundary conditions
By Liu-Renardy, Lasiecka-Triggiani
Thermoelastic plate equation is analytic in the Hilbert space setting. In the Hilbert space maximal regularity is generation of analytic semigroup is eqivalent.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 9 / 38
SLIDE 10
Purpose of my talk
To show Lp − Lq maximal regularity. We know Lp − Lq maximal regularity implies the generation of analytic semigroup. But it is not clear the opsite direction in the Banach space case. In fact, Kalton-Lancien proved that if every generator of an analytic on Banach space X has Lp maximal regularity, then X must be a Hilbert space.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 10 / 38
SLIDE 11
Setting function spaces.
I rewrite thermoelastic plate equation the following: Ut − AqU = F U|t=0 = U0, where I have set
Aq =
1
−∆2 −∆ ∆ ∆ ,
U =
u
∂tu θ ,
F =
f1 f2
,
U0 =
u0 u1
θ0
X = {F = (˜ f, f1, f2)|˜ f ∈ W2
q,D(Ω), f1 ∈ Lq(Ω), f2 ∈ Lq(Ω)},
E = {U = (u, v, θ)|u ∈ W4
q,D(Ω), v ∈ W2 q,D(Ω), θ ∈ W2 q,0(Ω)},
W2
q,0(Ω) = {u ∈ W2 q(Ω) | u|∂Ω = 0},
Wm
q,D(Ω) = {u ∈ Wm q (Ω) | u|∂Ω = ∂νu|∂Ω = 0}.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 11 / 38
SLIDE 12
The main theorem Ω: a bounded domain. ∂Ω:C4-hypersurface.
Let 1 < p, q < ∞. I assume there exist a γ0 > 0 such that U0, F satsify the following U0 ∈ [X, E]1−1/p,p, eγtF ∈ Lp(R+, Lq(Ω)), (γ ∈ [0, γ0]), Then there exists a unique solution: U ∈ Mp((0, ∞) : A) = Lp((0, T), E) ∩ W1
p((0, T), X)
satisfying the following estimates
∥eγt U∥Lp(R+,E) + ∥eγt Ut∥Lp(R+,X) ≦ Cp,q { ∥eγt F∥Lp(R+,Lq(Ω)) + ∥ U0∥[X,E]1−1/p,p } .
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 12 / 38
SLIDE 13
Outline of proof
To prove the theorem, I study two cases. One is the problem with zero right members and initial value.
∂2
t u + ∆2u + ∆θ = 0
in R+ × Ω,
∂tθ − ∆θ − ∆∂tu = 0
in R+ × Ω, u = ∂νu = θ = 0
- n R+ × ∂Ω
u(0, x) = u0(x), ∂tu(0, x) = u1(x), θ(0, x) = θ0(x) in Ω This case can be studied by analytic semigroup theory and interpolaion theory.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 13 / 38
SLIDE 14
Outline of proof
Another is the problem with non-zero right members and zero initial value. This case is the essential part in my proof.
∂2
t u + ∆2u + ∆θ = f1
in R+ × Ω,
∂tθ − ∆θ − ∆∂tu = f2
in R+ × Ω, u = ∂νu = θ = 0
- n R+ × ∂Ω,
u(0, x) = ∂tu(0, x) = θ(0, x) = 0 in Ω. In a Hilbert space case I know an elegant theory due to Lasiecka-Triggiani. But this theory can not be applied in the Banach space case. And I have to use exact solution formula in a whole and a half space, to get a priori estimate of the solution. Finally by the decomposition of unity, I localize the problem. Then I apply a whole and a half space case .
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 14 / 38
SLIDE 15
The generation of the analytic semigroup and its estimates A half space (N.-Shibata)
1 < q < ∞.
Aq generates the analytic semigroup : {Tq(t)}t≥0 on X. A bounded domain (Denk-Racke-Shibata ’08) Ω: a bounded domain, ∂Ω: C4-hypersurface. Let 1 < q < ∞. Aq generates the analytic semigroup : {Tq(t)}t≥0 on X and {Tq(t)}t≥0
satisfys the following estimates:
∃σ > 0, ∀t > 0, ∀G ∈ X(Ω), ∥Tq(t)G∥X ≤ Ce−σt∥G∥X.
where C is independent of t and F.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 15 / 38
SLIDE 16
A model problem ～a half space～ ∂2
t u + ∆2u + ∆θ = f1
in R × Rn
+,
∂tθ − ∆θ − ∆∂tu = f2
in R × Rn
+,
u = ∂xnu = θ = 0
- n R × ∂Rn
+,
u = ∂tu = θ = 0 in Rn
+.
By technical reason instead of this problem I consider the following problem which I shift with repect to time variable.
(∂t + 1)2u + ∆2u + ∆θ = f1
in R × Rn
+,
(∂t + 1)θ − ∆θ − (∂t + 1)∆u = f2
in R × Rn
+,
u = ∂xnu = θ = 0
- n R × ∂Rn
+,
u = ∂tu = θ = 0 in Rn
+.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 16 / 38
SLIDE 17
A model problem ～a half space～ (∂t + 1)2u + ∆2u + ∆θ = f1 + 2∂tu + u, (∂t + 1)θ − ∆θ − (∂t + 1)∆u = f2 + θ − ∆u
Lower order terms appear. When I consider the lower order terms, I use the estimates of the semigroup. Therefore I can consider this problem.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 17 / 38
SLIDE 18
To get the exact solution formula. The odd extension of (f1, f2), x = (x′, xn) x′ = (x1, x2, · · · , xn−1).
fo(t, x) =
f(t, x) xn > 0
−f(t, x′, −xn)
xn < 0 . I use the odd extension and it becomes a whole space problem.
(∂t + 1)2uw + ∆2uw + ∆θw = fo
1
in Rn+1,
(∂t + 1)θw − ∆θw − (∂t + 1)∆uw = fo
2
in Rn+1, I can estimate uw, θw by using the estimate of a whole space problem .
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 18 / 38
SLIDE 19
To get the exact solution formula.
Setting u = uw + w, θ = θw + τ, I consider this problem. Thank for uw, θw, right members become zero.
u = uw + w, θ = θw + τ. (∂t + 1)2w + ∆2w + ∆τ = 0
in R × Rn
+,
(∂t + 1)τ − ∆τ − (∂t + 1)∆w = 0
in R × Rn
+,
w = τ = 0, ∂xnw = M
- n R × ∂Rn
+,
Where I have set M = −∂xnuw.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 19 / 38
SLIDE 20
To get the exact solution formula. The partial Fourier-Laplace transform Lg(λ, ξ′, xn) =
- Rn e−λt−ix′·ξ′ g(t, x′, xn) dx′dt = ˆ
g,
L−1h(t, x′, xn) =
1
(2π)n
- Rn eλt+ix′·ξ′h(λ, ξ′, xn) dξ′dτ
λ = γ + iτ, x′ = (x1, x2, · · · , xn−1), ξ′ = (ξ1, ξ2, · · · , ξn−1) It becomes the ordinary equation with respect to xn. (λ + 1)2 ˆ
w(λ, ξ) + (∂2
xn − |ξ′|2)2 ˆ
w(λ, ξ) + (∂2
xn − |ξ′|2)ˆ
τ(λ, ξ) = 0, (λ + 1)ˆ τ(λ, ξ) − (∂2
xn − |ξ′|2)ˆ
τ(λ, ξ) − (λ + 1)(∂2
xn − |ξ′|2)ˆ
w(λ, ξ) = 0,
ˆ
w(λ, ξ′, 0) = ˆ
τ(λ, ξ′, 0) = 0, ∂xn ˆ
w(λ, ξ′, 0) = ˆ M(λ, ξ′, 0).
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 20 / 38
SLIDE 21
To get the exact solution formula.
Solving the ordinary equation, I have exact solution formula.
The displacement:u = uw + w, The temperature:θ = θw + τ.
w(t, x) =
3
∑
j=1
L−1 [ σj
det ∆(λ, ξ′) e−Aj(λ,ξ′)xn ˆ M(λ, ξ′, 0)
] (t, x′), τ(t, x) = −
3
∑
j=1
L−1 (λ + 1)(γ2
j + 2)σj
det ∆(λ, ξ′) e−Aj(λ,ξ′)xn ˆ M(λ, ξ′, 0)
(t, x′).
Aj =
√ (γj)−1(λ + 1) + |ξ′|2 (j = 1, 2, 3). :
The characteristic roots of the ordinary equation with respect to xn.
γ1 = α, γ2 = β, γ3 = ¯ β :
The characteristic roots of the ordinary equation with respect to t.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 21 / 38
SLIDE 22
The displacement:u = uw + w,the temperature:θ = θw + τ.
w(t, x) =
3
∑
j=1
L−1 [ σj
det ∆(λ, ξ′) e−Aj(λ,ξ′)xn ˆ M(λ, ξ′, 0)
] (t, x′), τ(t, x) = −
3
∑
j=1
L−1 (λ + 1)(γ2
j + 2)σj
det ∆(λ, ξ′) e−Aj(λ,ξ′)xn ˆ M(λ, ξ′, 0)
(t, x′). σ1 + σ2 + σ3 = 0. Lopatinski determinant
det ∆(λ, ξ′) = −[(γ2
2 − γ2 3)A1 + (γ2 3 − γ2 1)A2 + (γ2 1 − γ2 2)A3].
Aj =
√ (γj)−1(λ + 1) + |ξ′|2.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 22 / 38
SLIDE 23
A theorem due to Shibata-Shimizu
1 < p, q < ∞. We assume m(ξ′, λ) satisfies the multiplier condition that means the following:
|∂α′
ξ′ (τk∂k τ m(ξ′, λ))| ≦ Cα′(|λ|1/2 + |ξ′| + 1)−|α′|.
where we set α′ : multi-index, λ = γ + iτ with γ ≧ 0, ξ′ ∈ Rn−1 \ {0}, k = 0, 1. Then we set
[Kjg](t, x) = ∫ ∞ L−1 [
m(λ, ξ′)|ξ′|2 e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn) Aj − |ξ′|
ˆ
g(λ, ξ′, yn)
]
dyn Kjg satisfies the following:
∥e−γtKjg∥Lp(R,Lq(Rn
+)) ≦ Cp,q∥g∥Lp(R,Lq(Rn +)), j = 1, 2, 3. Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 23 / 38
SLIDE 24
The estimate of Lopatinski determinant.
To use the theorem due to Shibata-Shimizu, I have to estimate Lopatinski determinant.
Lopatinski determinant
det ∆(λ, ξ′) = −[(γ2
2 − γ2 3)A1 + (γ2 3 − γ2 1)A2 + (γ2 1 − γ2 2)A3].
To estimate the Lopatinski determinant, I divide the domain into two parts like these.
- 1
det ∆(λ, ξ′)
- ≤ C
1
|λ|
1 2 + |ξ′| + 1
when|λ + 1|
|ξ′|2 ≥ r/2,
- 1
det ∆(λ, ξ′)
- ≤ C
|ξ′| |λ + 1|
when|λ + 1|
|ξ′|2 ≤ r.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 24 / 38
SLIDE 25
The displacement:u = uw + w,the temperature:θ = θw + τ.
w(t, x) =
3
∑
j=1
L−1 [ σj
det ∆(λ, ξ′) e−Aj(λ,ξ′)xn ˆ M(λ, ξ′, 0)
] (t, x′), τ(t, x) = −
3
∑
j=1
L−1 (λ + 1)(γ2
j + 2)σj
det ∆(λ, ξ′) e−Aj(λ,ξ′)xn ˆ M(λ, ξ′, 0)
(t, x′).
When I estimate ∂4
nw for |λ + 1|
|ξ′|2 ≤ r,
the property that σ1 + σ2 + σ3 = 0 is a keypoint. Estimating othemr parts, I can estimate of w, τ. Moreover estimating uw, θw, I can get the maximal regularity in a half space.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 25 / 38
SLIDE 26
The maximal regularity in a half space.
Let 1 < p, q < ∞, F ∈ Lp,c(R, Lq(Rn
+)) = {u ∈ Lp(R, Lq(Rn +)) | u = 0 for t < 0}.
. Then there exists a unique solution : U ∈ Lp,c(R, E) ∩ W1
p,c(R, X)
satisfying the following estimates
∥U∥Lp(R,E) + ∥Ut∥Lp(R,X) ≦ Cp,q∥F∥Lp(R,Lq(Rn
+)). Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 26 / 38
SLIDE 27
Interior estimate
About the interior estimate I use a cutoff function, and it becomes a whole space problem.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 27 / 38
SLIDE 28
A priori estimate near boundary.
Let x0 be a point on the boundary. The neighborfood of x0
= ⇒
a bent half space
= ⇒
a half space
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 28 / 38
SLIDE 29
Reduction to a half space problem. The definition of a bent half space
Hω = {y = (y1, . . . , yn) ∈ Rn | yn > ω(y′), y′ ∈ Rn−1},
∂Hω = {y = (y1, . . . , yn) ∈ Rn | yn = ω(y′), y′ ∈ Rn−1}.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 29 / 38
SLIDE 30
Reduction to a half space problem.
I pick up x0 from the boundary of the domain. And I consider the innner normal vector to boundary at x0. Let vectors be B1 = t(B1
1, B1 2, · · · , B1 n), · · · , Bn−1 = t(Bn−1 1
, Bn−1
2
, · · · , Bn−1
n
)
such that they satisfy Bj · Bk = δjk, (j, k = 1, 2, · · · n). Set B = (B1, B2, · · · , Bn) ,B is an orthogonal matrix.
The first change of the variable
y = TB(x − x0)
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 30 / 38
SLIDE 31
Reduction to a half space problem. The unit outer vector of the origin after the change of the variable ˜ ν(0, · · · , 0) = T(0, . . . , 0, −1)
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 31 / 38
SLIDE 32
Reduction to a whole space problem.
By this change of the variable it becomes the following half space problem.
(∂t + 1)2z + ∆2z + ∆χ = H + R1
in R × Rn
+,
(∂t + 1)χ − ∆χ − (∂t + 1)∆z = I + R2
in R × Rn
+,
z = ∂nz = χ = 0
- n R × Rn
0,
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 32 / 38
SLIDE 33
The proof of the maximal regularity in a half space. The decomposition of w:w = w∞ + w0.
w∞(t, x) =
3
∑
j=1
L−1 [σje−Aj(λ,ξ′)xnψ∞(|λ+1|
|ξ′|2 )
det ∆(λ, ξ′)
ˆ
M(λ, ξ′, 0)
] (t, x′),
w0(t, x) =
3
∑
j=1
L−1 [σj(e−Aj(λ,ξ′)xn − e−|ξ′|xn)ψ0(|λ+1|
|ξ′|2 )
det ∆(λ, ξ′)
ˆ
M(λ, ξ′, 0)
] (t, x′).
Since σ1 + σ2 + σ3 = 0, I can add e−|ξ′|xn. In this talk I explain only w0(t, x).
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 33 / 38
SLIDE 34
The proof of the maximal regularity in a half space.
w0(t, x)
=
3
∑
j=1
L−1 [σj(e−Ajxn − e−|ξ′|xn)ψ0( |λ+1|
|ξ′|2 )
det ∆(λ, ξ′)
ˆ
M(λ, ξ′, 0)
] (t, x′), =
3
∑
j=1
∫ ∞ ∂ ∂yn L−1 [σj(e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn))ψ0
det ∆(λ, ξ′)
ˆ
M(λ, ξ′, yn)
]
dyn,
=
3
∑
j=1
∫ ∞ L−1 [σj(e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn))ψ0
det ∆(λ, ξ′)
∂n ˆ
M(λ, ξ′, yn)
]
dyn
+
3
∑
j=1
∫ ∞ L−1 [σj(−Aje−Aj(xn+yn) + |ξ′|e−|ξ′|(xn+yn))ψ0
det ∆(λ, ξ′)
ˆ
M(λ, ξ′, yn)
]
dyn
= w0
0(t, x) + w1 0(t, x).
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 34 / 38
SLIDE 35
The proof of the maximal regularity in a half space.
In this talk I estimate only ∂4
xnw0 0(t, x).
∂4
xnw0 0(t, x)
=
3
∑
j=1
∂4
xn
∫ ∞ L−1 [σje−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn)ψ0
det ∆(λ, ξ′)
∂n ˆ
M(λ, ξ′, yn))
]
dyn
=
3
∑
j=1
∫ ∞ L−1 [σj(A4
j − |ξ′|4)e−Aj(xn+yn)ψ0
det ∆
∂n ˆ
M(λ, ξ′, yn)
]
dyn
+
3
∑
j=1
∫ ∞ L−1 [σj|ξ′|4(e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn))ψ0
det ∆
∂n ˆ
M(λ, ξ′, yn)
]
dyn,
= w0,a + w0,b
0 .
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 35 / 38
SLIDE 36
The proof of the maximal regularity in a half space. λ + 1 = (Aj − |ξ′|)(Aj + |ξ′|)γj,
Aj =
√ (γj)−1(λ + 1) + |ξ′|2.
w0,b
0 (t, x)
=
3
∑
j=1
∫ ∞ L−1 [σj|ξ′|4(e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn))ψ0
det ∆
∂n ˆ
M(λ, ξ′, yn)
]
dyn
=
3
∑
j=1
∫ ∞ L−1 [ σj(λ + 1)|ξ′|4ψ0 γj(Aj + |ξ′|) det ∆
e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn) Aj − |ξ′|
∂n ˆ
M
]
dyn,
=
3
∑
j=1
∫ ∞ L−1 [
m(λ, ξ′)|ξ′|2 e−Aj(λ,ξ′)(xn+yn) − e−|ξ′|(xn+yn) Aj − |ξ′|
|ξ′|2∂n ˆ
M
]
dyn where I set m(λ, ξ′) =
(λ+1)ψ0( |λ+1|
|ξ′|2 )
γj((Aj+|ξ′|) det ∆.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 36 / 38
SLIDE 37
The proof of the maximal regularity in a half space.
- ∂α′
ξ′ m(ξ′, λ)
- =
- ∑
α′
1+α′ 2+α′ 3=α′
(λ + 1) { ∂α′
1ψ0
(|λ + 1| |ξ′|2 )}{ ∂α′
2
1
γj((Aj + |ξ′|) }{ ∂α′
3
1 det ∆
}
- ,
≤ ∑
α′
1+α′ 2+α′ 3=α′
|λ + 1|
1
(|λ|1/2 + |ξ′| + 1)|α′
1|
1
|ξ′|1+|α′
2|
|ξ′|1 |λ + 1||ξ′||α′
2| ,
≤ Cα′
1
(|λ|1/2 + |ξ′| + 1)|α′| .
Aj =
√ (γj)−1(λ + 1) + |ξ′|2,
- 1
det ∆(λ, ξ′)
- ≤ C
|ξ′| |λ + 1|
.
Yuka Naito (Sato) (Waseda University) thermoelastic plate equation 25/08/2009 37 / 38
SLIDE 38