Lp-Lq-Estimate of the Linear Equations of Thermoelasticity for Rhombic Media in R2

Universit¨at Konstanz

L ^p -L ^q -Estimate of the Linear Equations of Thermoelasticity for Rhombic Media in R ²

Monika Susanne Doll

Konstanzer Schriften in Mathematik und Informatik Nr. 210, September 2005

ISSN 1430–3558

c

Fachbereich Mathematik und Statistik c

Fachbereich Informatik und Informationswissenschaft Universit¨at Konstanz

Fach D 188, 78457 Konstanz, Germany Email: preprints@informatik.uni–konstanz.de

Konstanzer Online-Publikations-System (KOPS) URL: http://www.ub.uni-konstanz.de/kops/volltexte/2007/2230/

URN: http://nbn-resolving.de/urn:nbn:de:bsz:352-opus-22308

L

-L

-Estimate of the Linear Equations of Thermoelasticity for Rhombic Media in R

Monika Susanne Doll

Abstract: We determine theL^p-L^q-estimate of the linear equations of thermoelasticity for rhom- bic media in R². For this purpose we will transform the equations to an evolution equation showing that this equation has a unique solution with the help of the semigroup theory. In order to obtain the L^p-L^q-estimate, we will first apply the Fourier transformation to the evolution equation, thus determining the spectral representation of the solution. Then we will expand the eigenvalues of the Fourier transform operator using the Newton polygon procedure. The L^p-L^q-estimate is finally obtained by applying the method of stationary phase.

1 Introduction

Why do we studyL^p-L^q-estimates of the linear equations of thermoelasticity for rhombic media inR²? To answer this question we shall have a look back.

For differential operators the characteristic manifold of which has k vanishing main curva- tures at the most we obtain the decay rate (n−k)/2 for the dimension n. In the case of the equations of elasticity of isotropic media, for example, k = 1 for n = 3, the decay rate thus corresponding to 1.

In the case of the equations of elasticity of cubic media, unfortunately, k = 2 for n= 2. In order to obtain a decay of the solution also for this case we use the method of stationary phase ([5]). The decay rate depends on the order of the main curvatures’ vanishing. If the curvature does not vanish or if it vanishes of first or second order, respectively, we get ¹₂, ¹₃ or ¹₄ as decay rates. In the case of the equations of thermoelasticity of cubic media we get the decay rate ¹₂ ([1]) forn= 2 applying the method of stationary phase.

It was necessary to study another problem of this kind in order to find out whether either the isotropic case or the cubic case must be regarded as a special case. The calculation of the L^p-L^q-estimates for the linear equations of thermoelasticity of rhombic media showed that such calculations do not only become more and more sophisticated, but that the isotropic case proves to be the special case due to the decay rate ¹₂ found in it. We also have to assume that, applying the method of stationary phase, we cannot expect any better decay rates in the cases of the other anisotropic media.

In the following we consider the Cauchy problem for a homogeneous and initially rhombic medium in R². It is given by

u_tt− D⁰SDu+γ∇θ = 0, (1.1)

θ_t−∆θ+γ∇⁰u_t = 0 (1.2)

and the prescribed initial values

u(t= 0) =u₀, u_t(t= 0) =u₁, θ(t= 0) =θ₀, (1.3)

whereuandθare functions ofx∈R²andt≥0. The displacement vector of the two-dimensional medium is described byu=u(t, x), the temperature difference between the absolute temperature Tand the fixed reference temperatureT₀byθ=θ(t, x) =T(t, x)−T₀, and the coupling coefficient by γ. D is the formal generalized gradient

D:=





∂₁ 0 0 ∂₂

∂₂ ∂₁





and S:=





ν κ 0 κ ω 0

0 0 µ





contains the moduls of elasticity κ, µ, ν, ω ∈ R, which describe the rhombic medium. For physical reasons, S has to be positive. This means that

µ, ν, ω >0 (1.4)

and

νω > κ². (1.5)

As a result,

ν+ω−2|κ|>0. (1.6)

Using the semigroup theory we will now demonstrate that our initial value problem (1.1) to (1.3) has a unique solution. For this purpose we transform the linear equations of thermoelasticity formally to an evolution equation of first order in time and define

v=



 v¹ v² v³



:=



 SDu

u_t θ





as well as A:=





0 −SD 0

−D⁰ 0 γ∇ 0 γ∇⁰ −∆



.

Thus,v satisfies the differential equation

v_t+Av = 0 (1.7)

and the initial condition v(t= 0) =



 SDu₀

u1

θ₀



=:v0. (1.8)

As in section 2.1 in [3] we take the Hilbert space H:=L²(R²)

as a basis for our initial value problem and provide it with the weighted inner product (·,·)_H:= (·, Q·),

where Q:=





S⁻¹ 0 0 0 id 0

0 0 1





and (·,·) denotes the inner product in L²(R²). This means that the norm of the Hilbert space Hcorresponds to the energy. If we choose

D(A) :={v∈ H| v² ∈H¹(R²), v³∈H¹(R²), Av∈ H}, the following is valid:

Proposition 1.1 ([3], Theorem 2.2) −A is the generator of a contraction semigroup, and for each v₀∈D(A) there exists a unique solution

v∈C⁰([0,∞), D(A))∩C¹([0,∞),H) with v(t) =e^−tAv0.

In order to obtain the asymptotic behavior of the solution we decompose the solution implicitly as in the cubic case [1]. For this purpose we need the asymptotic expansion of the Fourier transform operator’s eigenvalues, which we determine with the help of the Newton polygon procedure. In order to obtain the asymptotic behavior of the solution we split up the solution due to the varying expansion of the eigenvalues. Here it is important whether the real parts of the eigenvalues disappear. The part depending on eigenvalues the real parts of which do not disappear at all can be estimated in a similar way as the curl-free part in the isotropic case [4]. To estimate the part depending on eigenvalues the real parts of which disappear we use the method of stationary phase. All in all we shall prove the same decay rates for rhombic media as for cubic media, which correspond to the decay rates of the divergence-free part of isotropic media:

|v(t)|q ≤c(1 +t)^{−12(1−2q)}kv₀kNp,p,

where 2≤q≤ ∞, ¹_p +¹_q = 1 as well asc >0 andN_p ∈Nare constants which depend onq only.

For the nonlinear equations of thermoelasticity of isotropic media in R³ Racke proved the existence of a unique, global and classical solution, decisively considering the decay rate of the temperature in his proof. Due to the lower decay rate of the temperature with rhombic media and especially with cubic media it is not possible to prove the existence of a unique, global and classical solution for the nonlinear equations of thermoelasticity of rhombic media in R², particularly not for cubic media inR².

2 Spectral Representation of the Operator

The first step to derive the L^p-L^q-estimate is the spectral representation of the operator A.

Applying the Fourier transformation to the equations (1.7) and (1.8), for eachξ = (ξ₁, ξ₂)∈R² we obtain

ˆ

v_t(t, ξ) + ˆA(ξ)ˆv(t, ξ) = 0, (2.1)

ˆ

v(t= 0, ξ) = ˆv0(ξ) (2.2)

mit

A(ξ) :=ˆ i

















0 0 0 −νξ₁ −κξ₂ 0

0 0 0 −κξ₁ −ωξ₂ 0

0 0 0 −µξ₂ −µξ₁ 0

−ξ₁ 0 −ξ₂ 0 0 γξ₁

0 −ξ₂ −ξ₁ 0 0 γξ₂

0 0 0 γξ₁ γξ₂ −i|ξ|²















 .

For the characteristic polynomialP₀ of ˆA(ξ) we calculate P0(ξ, λ) = det( ˆA(ξ)−λ id)

= λ

λ⁵− |ξ|²λ⁴+α|ξ|²λ³−β|ξ|⁴λ²+δ|ξ|⁴λ−η|ξ|⁶

=: λP(ξ, λ).

The meaning of α, β, δ and η is:

α ξ₁²

|ξ|², ξ²₂

|ξ|²

:= µ+ν+γ²− ξ₂²

|ξ|²(ν−ω), β

ξ₁²

|ξ|², ξ²₂

|ξ|²

:= µ+ν− ξ²₂

|ξ|²(ν−ω), δ

ξ₁²

|ξ|², ξ²₂

|ξ|²

:= µ(ν+γ²) +ξ₁²ξ₂²

|ξ|⁴ (ν−κ−2µ)(κ+ν+ 2γ²)−

−ξ₁²ξ₂²

|ξ|⁴ (ν−ω)(ν+γ²)− ξ₂⁴

|ξ|⁴(ν−ω)µ or

η ξ₁²

|ξ|², ξ₂²

|ξ|²

:= µν+ξ₁²ξ₂²

|ξ|⁴ (ν−κ−2µ)(κ+ν)−ξ₁²ξ₂²

|ξ|⁴ (ν−ω)ν−

− ξ₂⁴

|ξ|⁴(ν−ω)µ,

respectively. For allξ = (ξ₁, ξ₂)∈R² α, β, δ and η are positive functions according to [2].

The matrix ˆA(ξ) is diagonalizable for |ξ| 6= 0 under certain conditions regarding the moduls of elasticity κ, µ, ν, ω and the coupling coefficient γ. A detailed representation of the facts is provided by

Lemma 2.1 If κ, µ, ν, ω andγ satisfy the inequations ν ≥γ², ω ≥γ², νω−κ²

ν+ω−2κ ≥γ², νω−(κ+ 2µ)²

ν+ω−2(κ+ 2µ) ≥γ² (2.3)

as well as

(µ−ν−γ²)² ≥γ⁴, (µ−ω−γ²)² ≥ γ⁴, (κ+µ+γ²)² 4(µ−ν−γ²)(µ−ω−γ²)−4(κ+µ+γ²)²

(2µ−ν−ω−2γ²)²−4(κ+µ+γ²)² ≥ γ⁴ (2.4) and

µ+ν≥6γ², µ+ω ≥6γ², (2.5)

the eigenvalues of A(ξ)ˆ are different for |ξ| 6= 0, A(ξ)ˆ thus being diagonalizable for |ξ| 6= 0.

However, before starting to reflect on the actual proof we should note that we only have to consider the conditions

νω−(κ+ 2µ)²

ν+ω−2(κ+ 2µ) ≥γ² and

(κ+µ+γ²)² 4(µ−ν−γ²)(µ−ω−γ²)−4(κ+µ+γ²)² (2µ−ν−ω−2γ²)²−4(κ+µ+γ²)² ≥γ⁴, if

ν+ω−2κ−4µ6= 0 or

(2µ−ν−ω−2γ²)²−4(κ+µ+γ²)² 6= 0,

respectively, which is equivalent toν+ω−2κ−4µ6= 0 due to condition (1.6). Forν+ω−2κ−4µ= 0, both conditions have to be ignored completely.

We will prove Lemma 2.1 by showing that ˆA(ξ) does neither have a double real eigenvalue nor a double nonreal eigenvalue. For this purpose, however, we need a lower bound s for the real eigenvalues and the minimum m of the function h with h(Θ) = α²(Θ)−4δ(Θ) for the nonreal eigenvalues. The following is valid according to [2]:

s= min

( ν

ν+γ², ω

ω+γ², νω−κ²

νω−κ²+ (ν+ω−2κ)γ², νω−(κ+ 2µ)²

νω−(κ+ 2µ)²+h

ν+ω−2(κ+ 2µ)i γ²

)

|ξ|²

and

h(Θ) ≥ min (

(µ−ν−γ²)²,(µ−ω−γ²)²,

(κ+µ+γ²)²4(µ−ν−γ²)(µ−ω−γ²)−4(κ+µ+γ²)² (2µ−ν−ω−2γ²)²−4(κ+µ+γ²)²

)

=: m ≥ 0.

s > ¹₂|ξ|² and m ≥ γ⁴ due to the conditions (2.3) or (2.4), respectively. We shall now prove Lemma 2.1.

Proof:SinceP is a polynomial of odd degree and−Agenerates a contraction semigroup, ˆA(ξ) has at least one positive eigenvalue. If nonreal eigenvalues exist, these are conjugate-complex to each other as the coefficients of P are real. P is decomposed into linear factors overC:

P(ξ, λ) =

5

Y

j=1

(λ−λ_j(ξ)).

Thus

5

X

j=1

λ_j(ξ) =|ξ|²,

which means that the sum of the real parts of the eigenvalues of ˆA(ξ) equals |ξ|².

To show that ˆA(ξ) does not have a double real eigenvalue we use the lower bound s. We have chosen the conditions (2.3) such that s is always greater than ¹₂|ξ|². Suppose that ˆA(ξ) has a double real eigenvalue. Then the sum of the real parts of all eigenvalues is greater than|ξ|². Contradiction! For ˆA(ξ) does only have one positive real eigenvalue under the conditions (2.3).

To show that ˆA(ξ) does not have a double nonreal eigenvalue we use the minimummof the functionh. We have already chosen the conditions (2.4) such thatm is greater than or equal to γ⁴. Suppose thatλ=x+iy withy6= 0 is a double nonreal eigenvalue of ˆA(ξ). As each double zero ofP is generally also a zero of ^∂P_∂λ, λsolves the two equations

P(|ξ|,Θ, λ) = λ⁵− |ξ|²λ⁴+α(Θ)|ξ|²λ³−β(Θ)|ξ|⁴λ²+ +δ(Θ)|ξ|⁴λ−η(Θ)|ξ|⁶

= 0, (2.6)

∂P

∂λ(|ξ|,Θ, λ) = 5λ⁴−4|ξ|²λ³+ 3α(Θ)|ξ|²λ²−2β(Θ)|ξ|⁴λ+ +δ(Θ)|ξ|⁴

= 0. (2.7)

The following is valid for the imaginary part of equation (2.6) and for the real part of equation (2.7):

Im

P(|ξ|,Θ, x, y)

= yh

y⁴−10x²y²+ 5x⁴−4|ξ|²x(x²−y²)+

+α(Θ)|ξ|²(3x²−y²)−2β(Θ)|ξ|⁴x+

+δ(Θ)|ξ|⁴i

= 0, (2.8)

Re ∂P

∂λ(|ξ|,Θ, x, y)

= 5(x⁴−6x²y²+y⁴)−4|ξ|²x(x²−3y²) + + 3α(Θ)|ξ|²(x²−y²)−2β(Θ)|ξ|⁴x+ +δ(Θ)|ξ|⁴

= 0. (2.9)

Dividing equation (2.8) by −y and adding equation (2.9) to it, we obtain 4y⁴−20x²y²+ 8|ξ|²xy²−2α(Θ)|ξ|²y²= 0

or

y²= 5x²−2|ξ|²x+ 1

2α(Θ)|ξ|², (2.10)

respectively. We transform equation (2.8):

y⁴−

10x²−4|ξ|²x+α(Θ)|ξ|² y²+

+ 5x⁴−4|ξ|²x³+ 3α(Θ)|ξ|²x²−2β(Θ)|ξ|⁴x+δ(Θ)|ξ|⁴ = 0.

Together with equation (2.10) we get

y⁴= 5x⁴−4|ξ|²x³+ 3α(Θ)|ξ|²x²−2β(Θ)|ξ|⁴x+δ(Θ)|ξ|⁴. (2.11) Squaring equation (2.10) and inserting the result into equation (2.11), we obtain

0 = 20x⁴−16|ξ|²x³+ (4|ξ|⁴+ 2α(Θ)|ξ|²)x²−2γ²|ξ|⁴x+ +1

4α²(Θ)−δ(Θ)

|ξ|⁴

=

20x²−16|ξ|²x+ 2α(Θ)|ξ|² x²+ +

4x²−2γ²x+ 1

4α²(Θ)−δ(Θ)

|ξ|⁴

=: Ax²+B|ξ|⁴.

If we take B as a parabola inx, the functionB has its minimum in ¹₄γ² so that B ≥ −1

4γ⁴+1 4

α²(Θ)−4δ(Θ)

=−1 4γ⁴+ 1

4h(Θ).

h(Θ) ≥ γ⁴ and thus B ≥0 due to condition (2.4). We refute our assumption by showing that Ax²+B|ξ|⁴ >0 for all|ξ| 6= 0. Since the real eigenvalue ofP is greater than ¹₂|ξ|² and the sum of the real parts of the eigenvalues of P equals |ξ|², 0≤x < ¹₈|ξ|². B >0 for x= 0 and A >0 for 0< x≤ ¹₈α(Θ). Consequently,Ax²+B|ξ|⁴ >0, provided that 0≤x≤ ¹₈α(Θ). The following is valid for ¹₈α(Θ) < x < ¹₈|ξ|²:

A = 20

x−2 5|ξ|²

2

−16

5 |ξ|⁴+ 2α(Θ)|ξ|²

> 20 1

8|ξ|²− 2 5|ξ|²

2

− 16

5 |ξ|⁴+ 2α(Θ)|ξ|²

= −27

16|ξ|⁴+ 2α(Θ)|ξ|². Thus

Ax²+B|ξ|⁴ > −27

16|ξ|⁴x²+ 2α(Θ)|ξ|²x²+ 4|ξ|⁴x²−2γ²|ξ|⁴x+ +1

4

α²(Θ)−4δ(Θ)

|ξ|⁴

= 2α(Θ)|ξ|²x²+ 1

16|ξ|⁴(37x−32γ²)x+1

4h(Θ)|ξ|⁴

> 2α(Θ)|ξ|²x²+ 1 16|ξ|⁴

37

8 α(Θ)−32γ²

x+1

4h(Θ)|ξ|⁴,

which is, however, greater than or equal to zero if ³⁷₈α(Θ)−32γ²≥0 or the conditions (2.5) are

met. Contradiction! 2

If the conditions of Lemma 2.1 are met and ifPj(ξ) is the projector onto the eigenspace to the eigenvalue λ_j(ξ), the spectral decomposition of ˆA(ξ) is:

A(ξ) =ˆ

5

X

j=1

λ_j(ξ)P_j(ξ).

Hence, the solution ˆvof the transformed equations (2.1) and (2.2) can be represented as follows:

ˆ

v(t, ξ) =

5

X

j=1

e^−λj(ξ)tP_j(ξ)ˆv₀(ξ).

3 Eigenvalues of the Operator

From the proof of Lemma 2.1 we know that, under the conditions (2.3) to (2.5), ˆA(ξ) has one real eigenvalue and two pairs of conjugate-complex eigenvalues, where the real eigenvalue is positive and the real part of the nonreal eigenvalues not negative.

In the rhombic case we find the following decompositions for the polynomial P in the points Θ = 0, Θ = 1 and Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ:

P(|ξ|,Θ = 0, λ) = h

λ³− |ξ|²λ²+ (ν+γ²)|ξ|²λ−ν|ξ|⁴ih

λ²+µ|ξ|²i , P(|ξ|,Θ = 1, λ) = h

λ³− |ξ|²λ²+ (ω+γ²)|ξ|²λ−ω|ξ|⁴ih

λ²+µ|ξ|²i or

P

|ξ|,Θ = ν−κ−2µ ν+ω−2κ−4µ, λ

=

= (

λ³− |ξ|²λ²+ +

µ+1

2(ν+κ) +γ²−1 2

(ν−ω)(ν−κ−2µ) ν+ω−2κ−4µ

|ξ|²λ−

−

µ+1

2(ν+κ)−1 2

(ν−ω)(ν−κ−2µ) ν+ω−2κ−4µ

|ξ|⁴ )

·

·

λ²+ 1

2(ν−κ)−1 2

(ν−ω)(ν−κ−2µ) ν+ω−2κ−4µ

|ξ|²

,

respectively. Hence, we obtain purely imaginary eigenvalues for ˆA(ξ) in the points Θ = 0, Θ = 1 and Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ. The next lemma will show us that these points are the only points in which A(ξ) has purely imaginary eigenvalues.ˆ

Lemma 3.1 Let |ξ| 6= 0. In the case of (ν −κ−2µ)(ω −κ−2µ) > 0 the real part of the eigenvalues of A(ξ)ˆ disappears in the points Θ = 0, Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ and Θ = 1. In the case of (ν−κ−2µ)(ω−κ−2µ)≤0 with ν6=ω the real part of the eigenvalues of A(ξ)ˆ only disappears in the points Θ = 0 or Θ = 1.

Beweis:Letλ=iywithy6= 0. Then, the following equations result from the real and imaginary part of the polynomial P fory:

y⁴−β|ξ|²y²+η|ξ|⁴ = 0, (3.1)

y⁴−α|ξ|²y²+δ|ξ|⁴ = 0. (3.2)

Subtracting equation (3.2) from equation (3.1), we obtain (α−β)|ξ|²y²+ (η−δ)|ξ|⁴ = 0

or

y²=h

µ+ Θ(1−Θ)(ν+ω−2κ−4µ)i

|ξ|², (3.3)

respectively. If we insert equation (3.3) into equation (3.1), then

−Θ(1−Θ)h

Θ(ν+ω−2κ−4µ)−(ν−κ−2µ)i2

= 0. (3.4)

Equation (3.4) is now only valid if Θ = 0,

Θ = 1 or

Θ(ν+ω−2κ−4µ)−(ν−κ−2µ) = 0. (3.5)

In the case of ν+ω−2κ−4µ= 0, equation (3.5) is equivalent to ν−κ−2µ= 0.

For this reason, ω−κ−2µ= 0 and thus ν =ω=κ+ 2µ. In the case ofν+ω−2κ−4µ6= 0, Θ =˜ ν−κ−2µ

ν+ω−2κ−4µ (3.6)

solves equation (3.5). If ˜Θ≤0 or ˜Θ≥1, the following results from equation (3.6):

(ν−κ−2µ)(ω−κ−2µ)≤0,

and equation (3.4) is valid for Θ = 0 and Θ = 1 only. If 0< Θ˜ <1, the following results from equation (3.6):

(ν−κ−2µ)(ω−κ−2µ)>0,

and equation (3.4) is valid for Θ = 0, Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ and Θ = 1. 2 We shall now expand the eigenvalues of ˆA(ξ) for |ξ| →0 and |ξ| → ∞. SinceP is a polynomial of fifth degree, an explicit solution formula for its zeros does not exist. Nevertheless it is possible to expand the eigenvalues of ˆA(ξ) in power series with fractional powers at least for |ξ| → 0 and |ξ| → ∞. For this purpose we will keep Θ fixed from now on and apply a method based on Newton. Newton’s approach was to conceive ofP(|ξ|, λ) = 0 as an implicit equation forλand to calculate λusing an approximate procedure which gives an expansion ofλaccording to powers of |ξ|. If the conditions of the implicit function theorem are met,λcan be represented in|ξ| by a convergent power series. In general, however, this is not the case, which is also demonstrated by the following example:

P(|ξ|, λ) =|ξ|³−λ²= 0 namely has the solution

λ(|ξ|) =|ξ|^3/2.

This example already shows that, in any case, we need fractional powers of |ξ| to represent λ.

Using the following proposition proved in [2] we summarize the asymptotic expansions of the eigenvalues for |ξ| → 0 and |ξ| → ∞, numbering the eigenvalues as follows: In Θ = 0, let λ₁ be real, Reλ_2,3 > 0, Reλ_4,5 = 0 and Imλ_2,4 >0. Ifκ, µ, ν, ω and γ meet the requirements of Lemma 2.1, we have to distinguish the cases

(a) max{µ−ν, µ−ω} ≤0, (b) µ−ν≤0∧ µ−ω ≥2γ², (c) µ−ν ≥2γ² ∧ µ−ω ≤0, (d) min{µ−ν, µ−ω} ≥2γ²

due to the conditions (2.4). These conditions yieldκ+µ+γ² 6= 0 and therefore the additional cases

(i) κ+µ+γ² <0, (ii) κ+µ+γ²>0.

Proposition 3.2 The eigenvalues of A(ξ)ˆ can be expanded for |ξ| → 0 and |ξ| → ∞ in power series with fractional powers.

The expansions of the real eigenvalue λ₁ are

• for |ξ| →0 : λ1(|ξ|,Θ) = η(Θ)

δ(Θ)|ξ|²+O(|ξ|⁴),

• for |ξ| → ∞: λ₁(|ξ|,Θ) =|ξ|²−γ²+O(|ξ|⁻²).

The following is valid for the eigenvalues λ_2,3 and λ_4,5:

1. In the case of (ν −κ−2µ)(ω −κ−2µ) ≤ 0 with ν 6= ω the real part of the eigenvalues disappears in Θ = 0and Θ = 1.

(a) If max{µ−ν, µ−ω} ≤0, the expansions of the eigenvalues are as follows:

• for |ξ| →0 : λ_2/3(|ξ|,Θ) =±i

rα(Θ) 2 +1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰₁(Θ)>0 for all Θ∈[0,1],

λ_4/5(|ξ|,Θ) =±i

rα(Θ) 2 −1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰⁰₁(Θ) = 0 for Θ = 0 and Θ = 1,

• for |ξ| → ∞: λ_2/3(|ξ|,Θ) =±i

s β(Θ)

2η(Θ) − 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰₁(Θ)>0 for all Θ∈[0,1], λ_4/5(|ξ|,Θ) =±i

sβ(Θ)

2η(Θ) + 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰⁰₁(Θ) = 0 for Θ = 0 and Θ = 1.

(b) If µ−ν ≤0 and µ−ω≥2γ², the expansions of the eigenvalues are as follows:

• for |ξ| →0 : λ_2/3(|ξ|,Θ) =±i

rα(Θ) 2 +1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰₁(Θ) = 0 for Θ = 1,

λ_4/5(|ξ|,Θ) =±i

rα(Θ) 2 −1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰⁰₁(Θ) = 0 for Θ = 0,

• for |ξ| → ∞: λ_2/3(|ξ|,Θ) =±i

s β(Θ)

2η(Θ) − 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰₁(Θ) = 0 for Θ = 1, λ_4/5(|ξ|,Θ) =±i

sβ(Θ)

2η(Θ) + 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰⁰₁(Θ) = 0 for Θ = 0.

(c) If µ−ν ≥2γ² andµ−ω ≤0, the expansions of the eigenvalues are as follows:

• for |ξ| →0 : λ_2/3(|ξ|,Θ) =±i

rα(Θ) 2 −1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰₁(Θ) = 0 for Θ = 1,

λ_4/5(|ξ|,Θ) =±i

rα(Θ) 2 +1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰⁰₁(Θ) = 0 for Θ = 0,

• for |ξ| → ∞: λ_2/3(|ξ|,Θ) =±i

s β(Θ)

2η(Θ) + 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰₁(Θ) = 0 for Θ = 1, λ_4/5(|ξ|,Θ) =±i

s β(Θ)

2η(Θ) − 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰⁰₁(Θ) = 0 for Θ = 0.

(d) If min{µ−ν, ν−ω} ≥2γ², the expansions of the eigenvalues are as follows:

• for |ξ| →0 : λ_2/3(|ξ|,Θ) =±i

rα(Θ) 2 −1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰₁(Θ)>0 for all Θ∈[0,1],

λ_4/5(|ξ|,Θ) =±i

rα(Θ) 2 +1

2

pα²(Θ)−4δ(Θ)|ξ|+t⁰⁰₁(Θ)|ξ|²+O(|ξ|³), where t⁰⁰₁(Θ) = 0 for Θ = 0 and Θ = 1,

• for |ξ| → ∞: λ_2/3(|ξ|,Θ) =±i

s β(Θ)

2η(Θ) + 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰₁(Θ)>0 for all Θ∈[0,1], λ_4/5(|ξ|,Θ) =±i

s β(Θ)

2η(Θ) − 1 2η(Θ)

pβ²(Θ)−4η(Θ)

−1

|ξ|+ +t⁰⁰₁(Θ) +O(|ξ|⁻¹),

where t⁰⁰₁(Θ) = 0 for Θ = 0 uad Θ = 1.

2. In the case of (ν −κ−2µ)(ω −κ−2µ) > 0 the real part of the eigenvalues disappears in Θ = 0, Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ and Θ = 1.

(a) If max{µ−ν, µ−ω} ≤0, the expansions of the eigenvalues λ2,3 and λ4,5 have the same form as in 1 (a) for |ξ| →0 and |ξ| → ∞.

(i) For κ+µ+γ²<0 the coefficients satisfy:

t⁰₁(Θ) = 0 in Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ as well as t⁰⁰₁(Θ) = 0 in Θ = 0 and Θ = 1.

(ii) For κ+µ+γ² >0 the coefficients satisfy:

t⁰₁(Θ)>0 for allΘ∈[0,1]as well as

t⁰⁰₁(Θ) = 0 in Θ = 0, Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ andΘ = 1.

(b) If µ−ν≤0,µ−ω≥2γ², the expansions of the eigenvalues λ_2,3 and λ_4,5 have the same form as in 1 (b) for |ξ| →0 and |ξ| → ∞.

(i) For κ+µ+γ²<0 the coefficients satisfy:

t⁰₁(Θ) = 0 in Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ andΘ = 1 as well as t⁰⁰₁(Θ) = 0 in Θ = 0.

(ii) For κ+µ+γ² >0 the coefficients satisfy:

t⁰₁(Θ) = 0 in Θ = 1as well as

t⁰⁰₁(Θ) = 0 in Θ = 0 and Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ.

(c) If µ−ν≥2γ², µ−ω ≤0, the expansions of the eigenvalues λ_2,3 and λ_4,5 have the same form as in 1 (c) for |ξ| →0 and |ξ| → ∞.

(i) For κ+µ+γ²<0 the coefficients satisfy:

t⁰₁(Θ) = 0 in Θ = 1as well as

t⁰⁰₁(Θ) = 0 in Θ = 0 and Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ. (ii) For κ+µ+γ² >0 the coefficients satisfy:

t⁰₁(Θ) = 0 in Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ andΘ = 1 as well as t⁰⁰₁(Θ) = 0 in Θ = 0.

(d) Ifmin{µ−ν, µ−ω} ≥2γ², the expansions of the eigenvalues λ_2,3 andλ_4,5 have the same form as in 1 (d) for |ξ| →0 and |ξ| → ∞.

(i) Forr κ+µ+γ²<0 the coefficients satisfy:

t⁰₁(Θ)>0 for allΘ∈[0,1]as well as

t⁰⁰₁(Θ) = 0 in Θ = 0, Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ andΘ = 1.

(ii) For κ+µ+γ² >0 the coefficients satisfy:

t⁰₁(Θ) = 0 in Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ as well as t⁰⁰₁(Θ) = 0 in Θ = 0 and Θ = 1.

By expanding the eigenvalues it is possible to show the boundedness of the projectors P_j(ξ) =

5

Y

k=1k6=j

A(ξ)ˆ −λ_k(ξ)id λ_j(ξ)−λ_k(ξ) ,

in the zero point and at infinity, where j ∈ {1, . . . ,5}. Obviously, the following is valid for

|ξ| →0:

P_j(ξ) =O(1).

For|ξ| → ∞it can be also demonstrated in a similar way as used in [1] that P_j(ξ) =O(1).

4 Method of Stationary Phase

As we are now familiar with the expansion of eigenvalues in the zero point and at infinity, we are able to calculate the time-asymptotic behavior of solutionv of the differential equation (1.7) in relation to the initial condition (1.8) for infinite times. Recalling the representation of solution ˆ

v of the transformed equations (2.1) and (2.2) and performing the Fourier backtransformation we obtain for v:

v(t, x) = 1 2π

5

X

j=1

Z

R²

e^{i(xξ+iλj(ξ)t)}P_j(ξ)ˆv₀(ξ)dξ.

This means that the decay behavior of v is determined by the integrals Ij(t, x) :=

Z

R²

e^{i(xξ+iλj(ξ)t)}Pj(ξ)ˆv0(ξ)dξ,

wherej∈ {1, . . . ,5}. Let the subscriptj be fixed in the following.

In order to calculate the time-asymptotic behavior of I_j for t → ∞ we apply the method of stationary phase, focusing on the behavior of the phase

σ_jx t, ξ

:= x

tξ+iλ_j(ξ).

For our further reflections we parameterize ξ= (ξ₁, ξ₂)∈R² by polar coordinates:

ξ₁ := rcosϕ, ξ2 := rsinϕ,

wherer ≥0 and 0≤ϕ <2π, and define U_ε:=

(

0≤ϕ <2π

sin²ϕ≤ε ∨ 1−sin²ϕ≤ε ∨

ν−κ−2µ

ν+ω−2κ−4µ−sin²ϕ ≤ ε

2 )

as well as

R²_ε:={(ξ₁, ξ₂)∈R²| ξ₁=rcosϕ, ξ₂ =rsinϕ, wobeir ≥0, ϕ∈U_ε}. Since

Θ = ξ₂²

|ξ|² = sin²ϕ,

U_εis anε-neighborhood of Θ = 0, Θ = 1 and Θ = _ν+ω^ν−₋^κ_2κ^−2µ_−4µ. In the following we will determine the decay behavior of the integralI_j in dependence on its phase.

If the phase σ_j satisfies one of the two conditions (A) ∀ ξ ∈R² : Imσ_j(^x_t, ξ)6= 0,

(B) ∃ k₀, K₀ >0 ∃ C_j >0 ∃ ε₀ >0 ∀ 0< ε < ε₀ ∀ ξ∈R²

ε:

|∇ξσ_j(^x_t, ξ)| ≥C_j f¨ur|x| ≤ ¹₂k₀t oder |x| ≥2K₀t,

I_j decays in theL^p-L^q-estimate as (1+t)^−(1−2q). A detailed representation of the facts is provided by

Proposition 4.1 Let 2≤q≤ ∞ and ¹_p +¹_q = 1.

(a) IfN_p>3(1−²_q) (ifq ∈ {2,∞}, thenN_p≥3(1−²_q)) and the condition (A) is satisfied, then c=c(q)>0 exists. Thus, the following is valid for each v₀ ∈W^Np,p and t≥0 gilt:

|I_j(t)|_q≤c(1 +t)^−(1−2q)kv₀kNp,p.

(b) If N_p ≤3 and the condition (B) is satisfied, then c=c(q)>0 exists. Thus, the following is valid for each v₀ ∈W^Np,p and t≥0:

|I_j(t)|_q≤c(1 +t)^−(1−2q)kv₀kNp,p.

Proof: (a) We subdivide the proof into two steps. In the first step we determine the L²- L²-estimate

|I_j(t)|₂≤c|v₀|₂

forI_j and in the second step the special L¹-L^∞-estimate

|I_j(t)|_∞≤c(1 +t)⁻¹kv₀k3,1,

c >0 being a constant in both estimates. The claim finally results from the interpolation of the two estimates.

TheL²-L²-estimate is obvious. Since−Agenerates a contraction semigroup, the eigenvalues have a non-negative real part. Thus,

|I_j(t)|₂= 2π

e^−λjtP_jvˆ₀

2≤2π|e^−λjt|_∞|P_jvˆ₀|₂ ≤c|ˆv₀|2 = c|v₀|2, for the projector P_j is bounded.

We shall now prove the L¹-L^∞-estimate. Due to proposition 3.2 there are positive constants r, R, c_j andC_j so that for each|ξ| ≤r

Imσ_jx t, ξ

≥c_j|ξ|² and for each|ξ| ≥R

Imσ_jx t, ξ

≥C_j.

In the case of r <|ξ| < RImσ_j is also greater than zero. Since [r, R] is compact, ˜C_j >0 exists so that for each|ξ| ≤r

Imσ_jx t, ξ

≥c_j|ξ|² and for each|ξ|> r

Imσ_jx t, ξ

≥C˜_j. Thus we obtain

|Ij(t)|_∞ =

Z

R²

e^{iξ·−λj(ξ)t}Pj(ξ)ˆv0(ξ)dξ ∞

≤

|e^i·P_jˆv₀|_∞ Z

|ξ|≤r

e^−λj(ξ)t dξ+

+|e^i·(1 +| · |)³P_jvˆ₀|_∞ Z

|ξ|>r

e^−λj(ξ)t 1 (1 +|ξ|)³

dξ ∞

. We perform an upper estimate of the integral

Z

|ξ|≤r

e^−λj(ξ)tdξ

with the help of the transformation y=√ tξ:

Z

|ξ|≤r

e^−λj(ξ)tdξ≤ Z

|ξ|≤r

e^−cj|ξ|2tdξ= 1 t

Z

|y|≤r√ t

e^−cj|y|2dy.

This is the reason why fort≥1 Z

|ξ|≤r

e^−λj(ξ)tdξ≤ 1 t

Z

R²

e^−cj|y|2dy≤c(1 +t)⁻¹

and for 0≤t≤1 Z

|ξ|≤r

e^−λj(ξ)tdξ≤ 1 t

Z

|y|≤r√ t

dy≤πr².

Hence, the following is valid for all t Z

|ξ|≤r

e^−λj(ξ)tdξ≤c(1 +t)⁻¹.

The Integral Z

|ξ|>r

e^−λj(ξ)t 1 (1 +|ξ|)³ dξ

remains bounded due to the weight (1 +|ξ|)⁻³ and decays exponentially in time:

Z

|ξ|>r

e^−λj(ξ)t 1

(1 +|ξ|)³ dξ≤ce^−C˜jt.

However, we have to accept higher Sobolev norms in the estimate because of the weight:

|e^ix·(1 +| · |)³P_jˆv₀|_∞≤c

3

X

k=0

| · |^kP_jvˆ₀ ∞≤c

3

X

k=0

| · |^kvˆ₀

∞. We estimate each of the addends

| · |^kvˆ₀

∞

as follows:

| · |^kvˆ₀

∞ =

1 2π

Z

R²

| · |^ke^−ix·v₀(x)dx ∞

≤ c

k

X

l=0 2

X

m=1

Z

R²

∂^l

∂x^l_me^−ix·

v₀(x)dx ∞

≤ c

k

X

l=0 2

X

m=1

∂^l

∂x^l_mv₀ 1

.

Altogether we finally obtain the L¹-L^∞-estimate

|I_j(t)|_∞≤c(1 +t)⁻¹kv₀k3,1.

(b) We proceed analogously to (a). It is sufficient to show theL^∞-L^∞-estimate

|I_j(t)|_∞≤ct⁻¹

3

X

k=0

k | · |^kvˆ₀k_1,∞.

As the real part of the eigenvalue λ_j is not negative andP_j is bounded, the following is valid for each 0≤t≤1:

|Ij(t)|_∞= Z

R²

e^{i(ξ·+iλj(ξ)t)}Pj(ξ)ˆv0(ξ)dξ ∞

≤c Z

R²

|ˆv0(ξ)|dξ=c|vˆ0|1. In combination with Theorem 6.4 from [1] we obtain theL¹-L^∞-estimate

|I_j(t)|_∞≤c(1 +t)⁻¹kv₀k3,1

from the last two estimates, cbeing a positive constant in all estimates.

In order to prove the L^∞-L^∞-estimate, we split the integralI_j using the definition ofR²

ε: Ij(t, x) =

Z

R²ε

e^itσ(xt,ξ)Pj(ξ)ˆv0(ξ)dξ+ Z

R²\R²ε

e^itσ(xt,ξ)Pj(ξ)ˆv0(ξ)dξ.

The second integral can be estimated in the same way since the imaginary part of the phase σ_j differs from zero in R²\R²

ε. To obtain the time-asymptotic behavior of the first integral, we define

L:=

2

X

k=1

|∇ξσ|⁻²∂σ

∂ξ_k

∂

∂ξ_k

and the operator formally adjoint to L L^∗u:=−

2

X

k=1

∂

∂ξ_k

u|∇ξσ|⁻²∂σ

∂ξ_k

.

Since the projector P_j as well as all the derivatives of P_j are bounded and L e^itσ = ite^itσ, we obtain the following for the decay behavior in this case by partial integration:

Z

R²ε

e^itσ(xt,ξ)P_j(ξ)ˆv₀(ξ)dξ

=

1 it

Z

R²ε

Lh

e^itσ(xt,ξ)i

P_j(ξ)ˆv₀(ξ)dξ

= 1

t

Z

∂R²_ε

|∇ξσ|⁻²∂σ

∂νe^itσ(xt,ξ)P_j(ξ)ˆv₀(ξ)do_ξ−

− Z

R²ε

e^itσ(xt,ξ)L^∗P_j(ξ)ˆv₀(ξ)dξ

≤ c

t|(1 +| · |)²P_jvˆ₀|_∞ Z∞

0

1 r² dr+ +c

t|(1 +| · |)³L^∗P_jvˆ₀|_∞ Z∞

0

1 r²dr

≤ c t

3

X

k=0

k | · |^kP_jvˆ₀k_1,∞ ≤ c t

3

X

k=0

k | · |^kvˆ₀k_1,∞,

whereν is the outer unit normal at∂R²

ε. 2

Now we determine theL^p-L^q-estimate forIjin the case of ¹₂k0t <|x|<2K0t. For this purpose we introduce a transformationT which is similar to the transformation used for polar coordinates.

Instead of integrating over circles we integrate over the curve Γ_y(ϕ) := Γ(y, ϕ),

wherey >0,ϕ∈U_ε and Imλ_j(Γ_y(ϕ)) =y.

We define the transformation T by

T: (0,∞)×U_ε→(0,∞)×U_ε, (y, ϕ)7→(r, ϕ).

In order forT to be bijective and |det T⁰|<∞, we choose ε >0 such that 1

2Imλ_j(r,ϕ)˜ ≤Imλ_j(r, ϕ) ≤2 Imλ_j(r,ϕ)˜ and

1 2

∂

∂rImλ_j(r,ϕ)˜ ≤ ∂

∂rImλ_j(r, ϕ)≤2 ∂

∂rImλ_j(r,ϕ),˜ if

(sin²ϕ˜= 0 ∧ sin²ϕ≤ε) ∨ (sin²ϕ˜= 1 ∧ 1−sin²ϕ≤ε) ∨

sin²ϕ˜= ν−κ−2µ ν+ω−2κ−4µ ∧

ν−κ−2µ

ν+ω−2κ−4µ−sin²ϕ ≤ ε

2

. As an abbreviation we define

s :=

x t

, ζ := 1

s x

t

and

Φ_j(y, ϕ, ζ) := ζ





r(y, ϕ) cosϕ r(y, ϕ) sinϕ

−y+ix_j(y, ϕ)



, where

x_j := Reλ_j

and ζ is an element from the set M :=

( ζ ∈S²

ζ₃ >0, k₀

pk₀²+ 4 ≤ |(ζ₁, ζ₂)| ≤ 2K₀ p4K₀²+ 1

) . We further define

SP_ε^j :=

(y, ϕ, ζ)∈(0,∞)×Uε×M

∂Φ_j

∂ϕ (y, ϕ, ζ) = 0

.