Fredholm theory and spectral analysis

has a solution w = (w, ζ)^⊤ ∈Yη Since (sI− L)u= r∈ X_η¹ we obtain using integration by parts for all ϕ∈C₀^∞(R,R)

Z

R

u_xx(x)ϕ_x(x)dx= Z

R

A⁻¹[su−cu_x−S_ωu−Df(v_⋆)u−r](x)ϕ_x(x)dx

=− Z

R

A⁻¹[sux−cuxx−Sωux−Df(v⋆)ux−D²f(v⋆)[v⋆,x, u]−rx](x)ϕ(x)dx.

Thus ux∈ H_η¹∩H_loc² with

uxxx=A⁻¹[sux−cuxx−Sωux−Df(v⋆)ux−D²f(v⋆)[v⋆,x, u]−rx]∈L²_η. Therefore, u_x∈H_η² solves

(sI− L) ux

0

=

(sI −L)ux

0

=

∂_x[su−Au_xx−cu_x−S_ωu−Df(v_⋆)u] +D²f(v_⋆)[v_⋆,x, u]

0

=

rx−D²f(v⋆)[v⋆,x, u]

0

.

Since wis the unique solution of (3.37) we concludew=ux,ζ = 0. This proves i). Now ii) follows by (3.36) and by applying Lemma 3.10 to the equation

(sI − L) ux

0

=

rx−D²f(v⋆)[v⋆,x, u]

0

.

3.3. THE LINEARIZED OPERATOR L 73 In addition, let us consider the operator

Lη :H² →L², u7→ηLη⁻¹u.

A straightforward calculation shows thatLη can be written as a second order differential operator on L²

Lηu=Auxx+Bµux+Cµu with coefficients given by

Bµ(x) =cI + 2µx

√x²+ 1A,

Cµ(x) =Sω+Df(v⋆(x)) +µ²A x²

x²+ 1 −cµI x

√x²+ 1 −µA 1

√x²+ 1 − x² (x²+ 1)³²

! . The next step is to introduce the map

ψ :Xη →L²_η ×R², u

ρ

7→

u−ρˆv ρ

.

Taking the norm k(u, ρ)k²_L²_η×R :=|ρ|²+kuk²_L²_η onL²_η×R², it follows immediately that ψ defines an isometry fromXη to L²_η×R² and its inverse is given by

ψ⁻¹ :L²_η×R² →Xη, u

ρ

7→

u+ρˆv ρ

.

Lemma 3.12. The map ψ : Xη →L²_η ×R² is an isometric isomorphism. Moreover, if 0≤µ <2, then ψ :Yη →H_η²×R² is a homeomorphism.

Proof. By the previous observation it remains to show the continuity of ψ from Y_η to H_η²×R². From Proposition 2.7 we obtain for 0≤µ <2 and u= (u, ρ)^⊤ ∈Yη

kψ(u)k²H_η²×R² =|ρ|²+ X2

α=0

k∂^α(u−ρˆv)k²L²_η

≤(1 +kvˆxk²L²η +kvˆxxk²L²η)|ρ|²+ku−ρˆvk²L²η +kuxk²L²η+kuxxk²L²η ≤Ckuk²Yη.

With the homeomorphism ψ we can define the operator L^ψ :H_η²×R² →L²_η×R², u7→ψLψ⁻¹u

Again, a straightforward calculation shows that for u = (u, ρ)^⊤ the operator L^ψ can be written as

Lψu=

Auxx+cux+Sωu+Df(v⋆)u+Aρˆvxx +cρˆvx+ (Df(v⋆)−Df(v_∞))ρˆv S_ωρ+Df(v_∞)ρ

. If Assumption 1 and 2 are satisfied, it follows thatLψ defines a closed, linear operator on L²_η×R² withD(L^ψ) =H_η²×R². Furthermore, sinceψ is a homeomorphism and therefore a Fredholm operator of index 0we conclude from Lemma A.2 that the Fredholm indices ofsI−LandsI−Lψ coincide. The same holds true for the operatorssI−LandsI−L_η since the multiplication operator associated withηfrom Lemma 3.1 is a homeomorphism.

Furthermore, a compact perturbation argument will show that the variable coefficient operatorsI−L_η onL² has the same Fredholm index as the piecewise constant coefficient operator given by

Lη,∞ :H²→L², u7→Auxx+Bµ,∞ux+Cµ,∞u, where

Bµ,∞(x) =

(cI + 2µA, x ≥0

cI −2µA, x <0, Cµ,∞(x) =

(S_ω+Df(v_∞) +µ²A−cµI, x≥0 Sω+Df(0) +µ²A+cµI, x <0.

(3.39) We note and prove these observations in the following lemma.

Lemma 3.13. Let Assumption 1 and 2 be satisfied, 0≤µ≤ min(µ⋆,2) and s ∈C with µ⋆ from Theorem 2.6. Then the following statements are equivalent:

i) The operator (sI− L) :Y_η →X_η is a Fredholm operator of index k.

ii) The operator (sI− Lψ) :H_η²×R² →L²_η ×R² is a Fredholm operator of index k.

iii) The operator (sI−L) :H_η² →L²_η is a Fredholm operator of index k.

iv) The operator (sI−Lη) :H² →L² is a Fredholm operator of index k.

v) The operator (sI−Lη,∞) :H²→L² is a Fredholm operator of index k.

Proof. i) ⇔ii): By Lemma 3.12, the maps ψ :Xη →L²_η×R² and ψ :Yη →H_η²×R² are homeomorphisms and therefore Fredholm operators of index0. Thus, the equivalence of i) and ii) follows by Lemma A.2.

iii)⇔iv): By Lemma 3.1, the multiplication operatorsmη :L²_η →L² andmη :H_η² →H² are homeomorphisms and therefore Fredholm operators of index0. Thus, the equivalence

3.3. THE LINEARIZED OPERATOR L 75 of iii) and iv) follows by Lemma A.2.

ii) ⇔ iii): The operatorLψ can be decomposed into Lψ = ˜L+K where L˜is given by L˜:H_η²×R² →L²_η ×R², L˜

u ρ

:=

Auxx+cux+Sωu+Df(v⋆)u (Sω+Df(v_∞))ρ

and the operator K by

K :H_η²×R² →L²_η ×R², K u

ρ

:=

Aρˆv_xx+cρˆv_x+ (Df(v_⋆)−Df(v_∞))ρˆv 0

. Since sI −S_ω−Df(v_∞) ∈ R^2,2 is a Fredholm operator of index 0 on R², Lemma A.3 implies that sI−L˜is a Fredholm operator of index k if and only if (sI−L) :H_η² →L²_η is. We show that K is a compact operator. Then the assertion follows by Lemma A.4.

To see the compactness of K, let {u_n}n∈N ⊂ H_η² ×R², u_n = (u_n, ρ_n)^⊤ be a bounded sequence and let C > 0 denote a universal constant. Then there exists a subsequence ρnk such that ρnk →ρ as k→ ∞. We define

w:=Aρˆvxx+cρˆvx+ (Df(v⋆)−Df(v_∞))ρˆv.

Then Assumption 1 and Theorem 2.6 imply w∈L²_η. Moreover, we have k(Df(v⋆)−Df(v_∞))(ρnk −ρ)ˆvkL²η

≤ k(Df(v⋆)−Df(v_∞))(ρnk −ρ)ˆvkL²η(R₋)+k(Df(v⋆)−Df(v_∞))(ρnk−ρ)ˆvkL²η(R₊)

≤CkˆvkL²η(R₋)|ρnk−ρ|+C|ρnk −ρ|kDf(v⋆)−Df(v_∞)kL²η(R₊)

≤Ckˆvk^L2η(R−)|ρnk−ρ|+C|ρnk −ρ|kv⋆ −v_∞k^L2η(R₊)≤C|ρnk−ρ|. This implies for w= (w, ρ)^⊤∈L²_η ×R²

kKu_n

k−wkL²η×R² ≤ |ρ_nk −ρ|+kAρ_nkvˆ_xx+cρ_nkvˆ_x+ (Df(v_⋆)−Df(v_∞))ρ_nkvˆ−wkL²η

=|ρnk−ρ|+kA(ρnk−ρ)ˆvxx+c(ρnk −ρ)ˆvx+ (Df(v⋆)−Df(v_∞))(ρnk −ρ)ˆvkL²η

≤C(1 +kvˆxxk^L2η+kˆvxk^L2η)|ρnk−ρ|+k(Df(v⋆)−Df(v_∞))(ρnk −ρ)ˆvk^L2η

≤C|ρnk −ρ| →0, k→0.

Thus, Ku_n

k → w in L²_η ×R² as k → ∞. This shows the compactness of K and the assertion follows by Lemma A.4.

iv) ⇔ v): Since the operator ∂x : H^k+1 → H^k, k ≥ 0 is bounded, Lemma D.4 and Theorem 2.6 imply that the operator

Lη −Lη,∞ :H² →L², u7→(Bµ−Bµ,∞)ux+ (Cµ−Cµ,∞)u is compact. Hence the assertion follows by Lemma A.4.

The lemma shows that the Fredholm properties ofsI − L on Xη are determined by the piecewise constant coefficient operator sI−Lη,∞. Thus, we are interested into the solvabilty of the resolvent equation

(sI−Lη,∞)u=r, u∈H², r∈L², s∈C. (3.40) We use the classical approach, for instance, from [36] or [56], [32], where the solution of (3.40) is constructed using exponential dichotomies. For this purpose we transform (3.40) into a first order system via w= (u, u^′)and obtain

M(s)w=h, M(s) =∂_x−M(s,·), h= (0, r)^⊤ (3.41) with

M(s, x) =

(M₊(s), x≥0

M₋(s), x <0, M_±(s) =

0 I A⁻¹(sI −C_±) −A⁻¹B_±

and the matrices B_±, C_± given by

B_± :=cI ∓2µA, C_± :=Sω+Df(v_±) +µ²A∓cµI.

From [22] we have that the operatorM(s)has an exponential dichotomy on the half-line R± if and only if the matrix M_±(s) is hyperbolic, cf. Proposition B.4. Therefore, we define the set

ΩF :={s∈C:M+(s)and M₋(s) are hyperbolic}.

For s ∈ Ω_F we denote by m^±_s,u(s) the dimensions of the stable and unstable subspaces of M_±(s), i.e. m^±_s(s) denotes the sum of multiplicities of the eigenvalues of M_±(s) with negative real part and m^±_u(s) those with positive real part. Now we have the following classical result which can be found in several texts from the literature. See for instance [36, Lem. 3.1.10] or [48], [49], [56, Sec. 3].

Lemma 3.14. Let Assumption 1 and 2 be satisfied, 0 ≤ µ ≤ min(µ_⋆,2) with µ_⋆ from Theorem 2.6. Then the operatorsI−Lη,∞ :H² →L² is a Fredholm operator if and only if s ∈ΩF. If s∈ΩF then the Fredholm index is given by

ind(sI−Lη,∞) =m⁺_s (s)−m⁻_s(s).

Lemma 3.14 together with Lemma 3.13 imply that sI− L : Yη →Xη is a Fredholm operator if and only ifs∈Ω_F. Moreover, since the matricesM_± depend continuously on s ∈ C we conclude that the Fredholm index of sI − L stays constant in any connected component of ΩF. Recall the dispersion set from (0.27) given by

σdisp,µ(L) =σ_disp,µ⁻ (L)∪σ_disp,µ⁺ (L) Then we have the following lemma.

3.3. THE LINEARIZED OPERATOR L 77 Lemma 3.15. Let Assumption 1 and 2 be satisfied. Then the set ΩF is the complement of the dispersion set, i.e. ΩF =C\σdisp,µ(L).

Proof. First we show that det(−ν²A+iνB_±+C_±−sI) = 0 if and only ifM_±(s) is not hyperbolic. Assume M_±(s) is not hyperbolic, i.e. there exists ν ∈ R, w ∈ C⁴, |w| = 1 such that iνw =M_±(s)w. This implies, with w = (w1, w2)^⊤, that iνw1 =w2 and hence w1 6= 0 due to |w|= 1. Moreover,

−ν²w1 =A⁻¹(sI−C_±)w1−A⁻¹B_±w2 =A⁻¹(sI−C_±)w1−iνA⁻¹B_±w1. Hence,

(−ν²A+iνB_±+C_±)w₁ = 0.

Thus, det(−ν²A+iνB_±+C_±−sI) = 0.

Conversely, suppose det(−ν²A+iνB_±+C_±−sI) = 0. Then there exists w1 ∈C^m such that (−ν²A+iνB_±+C_±−sI)w₁ = 0. Now setting w₂=iνw₁ leads to

0 I A⁻¹(sI−C^±) −A⁻¹B^±

w1

w₂

=

w2

A⁻¹(sI −C_±)w₁−A⁻¹B_±w₂

=

iνw1

A⁻¹(sI −C_±)w1−iνA⁻¹B_±w1

=

iνw1

A⁻¹(sI−C_±−iνB_±)w1

=

iνw₁

−A⁻¹ν²Aw1

=iν w₁

w2

.

It holds true that the Fredholm index of sI − L stays constant in any connected component of ΩF, see [36]. Therefore, we are interested in the shape and location of the dispersion set and in particular in the connected components of ΩF. The Fredholm region Ω_F can be written as

ΩF =

s∈C: det(sI−D_±(ν))6= 0∀ν ∈R , D_±(ν) := −ν²A+iνB_±+C_±. Hence we look for eigenvalues of the matrixD_±(ν),ν ∈R. Its characteristic polynomial is given by

d^±(s) =s²−trD_±(ν)s+ detD_±(ν).

The roots of d^±(·, ν) can be computed explicitly. We have s∈σ_disp,µ⁺ (L) if and only if s= trD₊(ν)

2 ±

r(trD₊(ν))²

4 −detD₊(ν)

=−α1ν²+i(c−2α1µ)ν+µ²α1−cµ+g₁^′(|v_∞|²)|v_∞|²

±h

−α₂²ν⁴−4iα²₂µν³+ (6α₂²µ²+ 2α2g₂^′(|v_∞|²)|v_∞|²)ν² + 4i(α²₂µ³+µα2g₂^′(|v_∞|²)|v_∞|²)ν

−α²₂µ⁴−2α2µ²g₂^′(|v_∞|²)|v_∞|²+ (g^′₁(|v_∞|²)|v_∞|²)²i¹₂

(3.42)

and s ∈σ⁻_disp,µ(L) if and only if s= trD₋(ν)

2 ±

r(trD₋(ν))²

4 −detD₋(ν)

=−α1ν²+i(c+ 2α1µ)ν+µ²α1+cµ+g1(0)

±h

−α²₂ν⁴+ 4iα²₂µν³ + (6α²₂µ²+ 2α₂(g₂(0) +ω))ν²

−4iα₂(α₂µ³+µ(g₂(0) +ω))ν−α²₂µ⁴−2(g₂(0) +ω)α₂µ²−(g₂(0) +ω)²i¹₂ . (3.43) Roughly speaking, these are four curves in the complex plane running from −∞ −i∞ to −∞+i∞. In the special case α2 = 0 the equations (3.42) and (3.43) simplify to

s=−α₁ν²+i(c−2α₁µ)ν+µ²α₁−cµ+g₁^′(|v_∞|²)|v_∞|²±g₁^′(|v_∞|²)|v_∞|² and

s=−α1ν²+i(c+ 2α1µ)ν+µ²α1+cµ+g1(0)±i(g2(0) +ω).

Then the dispersion set consists of four parabolas in the complex plane opened to the left, see 0.4 b).

We are now in the position to formulate and prove the main result of this section de-scribing the essential spectrum of L on the exponentially weighted space Xη.

Theorem 3.16. Let the Assumption 1-3 be satisfied and 0 < µ < min(µ_ess, µ_⋆,2) with µess from Assumption 3 and µ⋆ from Theorem 2.6. Then there are ε > 0, γ <0 and a unique connected component Ω_∞ of ΩF satisfying:

i) Sε,γ :=

s ∈C:|arg(s−γ)|< ^π₂ +ε, s 6=γ ⊂Ω_∞.

ii) For all s∈Ω_∞ the operator sI− L:Yη →Xη is Fredholm of index 0.

3.3. THE LINEARIZED OPERATOR L 79 iii) ∂Ω_∞⊂σdisp,µ(L).

iv) σ_ess(L)⊂C\Ω_∞.

Proof. i). Fors ∈ΩF Lemma 3.13 and Lemma 3.14 imply the operatorsI − Lto be a Fredholm operator of index ind(sI− L) = m⁺_s (s)−m⁻_s (s). For s ∈ σdisp,µ(L) we have Res → −∞ as |s| → ∞. Thus, Reσdisp,µ(L)< ∞. Now let s0 >Reσdisp,µ(L). Recall for a matrix M ∈ C^m,m its lower spectral bound α(M) := min{Re (x^HMx) : |x| = 1} and let s0 be so large such that

α(s0I−C_±)≥s0−max{Re (v^HC_±v) :|v|= 1}> µ|α2| α₁ . Then for all s ≥s0 we also have

α(sI−C_±)≥s−max{Re (x^HC_±x) :|x|= 1}> µ|α2| α1

and since α(A) =α1 we obtain

|B_±−B_±^⊤|=|2µ(A−A^⊤)|= 4µ|α2|= 4α(A)µ|α₂| α1

<4α(A)α(sI−C_±) ∀s ∈[s0,∞).

Now for all s ∈ [s0,∞) Lemma D.1 implies M_±(s) to be hyperbolic with m^±_s (s) = 2.

Thus sI− L is a Fredholm operator of index 0. Since both M_± depend continuously on s we conclude that m^±_s (s) = 2 for s in the connected component Ω_∞ of ΩF containing [s0,∞). Thus sI − L is Fredholm index 0 for s ∈ Ω_∞ and ∂Ω_∞ ⊂ ∂ΩF = σdisp,µ(L).

Therefore ii) and iii) hold. Moreover, iv) follows by definition of the essential spectrum, cf. Definition 1.10. It remains to show i). Using ii) it is sufficient to show there is a sector Sε,γ, ε > 0, γ < 0 with σdisp,µ(L)∩ Sε,γ = ∅. The dispersion set consists of four curves given by the equations (3.42) and (3.43). For each of those we can choose a parametrizationχ:R→Csuch that (3.42), (3.43) respectively, is equivalent toχ(ν) =s and χ is given by

χ(ν) =−α1ν²+iξ1ν+ξ2± q

−α²₂ν⁴+p(ν), ν∈ R,

where ξ1, ξ2 ∈R and p(ν) =iβ3ν³+β2ν²+iβ1ν+β0 is a polynomial of degree 3over C with βi ∈R. For the derivative of the parametrization there holds

χ^′(ν) =−2α1ν+iξ1∓ 2α₂²ν³

p−α²₂ν⁴+p(ν) ± p^′(ν) p−α₂²ν⁴+p(ν). Now,

2α²₂ν²

p−α₂²ν⁴+p(ν) = 2α²₂

p−α²₂+ν⁻⁴p(ν)

ν→±∞

−→ 2α²₂

p−α₂² = 2α₂²

i|α2| =−2i|α2|.

Therefore,

ν⁻¹χ^′(ν)→ −2α1±2i|α2|, ν→ ±∞.

Let Tχ(ν) := _|^χ_χ^′′(ν)_(ν)| be the tangent vector of the curve atχ(ν). Then for ν >0 Tχ(ν) = ν⁻¹χ^′(ν)

|ν⁻¹χ^′(ν)| → −α1±i|α2|

|α| , ν → ∞. For ν <0 we obtain

Tχ(ν) =− ν⁻¹χ^′(ν)

|ν⁻¹χ^′(ν)| → α1∓i|α2|

|α| , ν → −∞.

˜ γ

˜ σ_disp,µ(L) ε

Sε,˜˜γ

a)

σ_disp,µ(L)

−β₀

Sε,˜˜γ

Sε,γ

b)

Figure 3.2: Geometric situation in the proof of Theorem 3.16.

SinceReχ(ν)→ −∞as ν → ±∞, we find a sector Sε,˜˜γ,γ >˜ 0, 0<ε <˜ tan⁻¹

α1

|α2|

such that σ_disp,µ(L)⊂(Sε,˜˜γ)^c, cf. Figure 3.2 a). Now Assumption 3 implies σ_disp,µ(L)⊂ {Res <−β0}. Then for every−β0 < γ <0there is0< ε≤ε˜such thatS^ε,γ∩σdisp,µ(L) =

∅, cf. Figure 3.2 b).

Remark 3.17. To fully describe the essential spectrum, according to Definition 1.10, of the linearized operator L one can use Lemma 3.14 and compute m⁺_s (s), m⁻_u(s) in the connected components ofΩF. In the connected components the dimensionsm⁺_s (s), m⁻_u(s) stay constant. The dimensions are given by the number of eigenvalues with negative real part of the matrices M+(s), M₋(s). The Fredholm index in then given by ind(sI− L) =

3.3. THE LINEARIZED OPERATOR L 81 m⁺_s (s)−m⁻_u(s). The essential spectrum is shown in Figure 3.3 in case of (QCGL) with parameters α = 1, µ=−¹₈, β = 1 +i, γ =−1 +i In this case the matrices are explicitly given by

M₊(s) =







0 0 1 0

0 0 0 1

s−2g₁^′(|v_∞|²)|v_∞|² 0 −c 0

−2g₂^′(|v_∞|²)|v_∞|² s 0 −c







and

M₋(s) =







0 0 1 0

0 0 0 1

s−g₁(0) −ω−g₂(0) −c 0 ω+g2(0) s−g1(0) 0 −c





.

The numbers in the connected components indicate the Fredholm index of sI − L. We

-2 -1 0

-2 0

2

⁻¹

−1

+1 +1

+1

+2

Figure 3.3: The essential spectrum of the linearized op-erator L (green) with the dispersion sets (red/blue) in an example of (QCGL). The numbers in the connected com-ponent of ΩF indicate the Fredholm indiex of sI− L.

note that the essential spectrum strongly depends in the choice of its definition which differs in the literature, cf. [25]. However, for the stability behavior of the TOF, the choice of the precise definition has no effects since the essential spectrum is bounded by the dispersion set in any case.

Now we take Assumption 4 into account and conclude the section by studying the zero eigenvalue. Assumption 4 states that zero is an eigenvalue of the linearized operator Lwith algebraic multiplicity at most2. Using the fact that the whole group orbita(γ)v_⋆, γ ∈ G is a stationary solution of the Cauchy problem (0.22) and that the group action a(γ) is continuously differentiable, we will see that one finds two linearly independent eigenfunctions of L. Then by Assumption 4 it follows that zero is an eigenvalue of algebraic multiplicity equal to 2.

Lemma 3.18. Let the Assumption 1-4 be satisfied and 0< µ <min(µess, µ⋆,2)with µess

from Assumption 3 and µ⋆ from Theorem 2.6. Then s = 0 is an eigenvalue of L with algebraic multiplicity two and linearly independent eigenfunctions given by

ϕ₁ =−S₁v_⋆ =−

S₁v_⋆ S1v_∞

∈Y_η, ϕ₂ =−v_⋆ =− v_⋆,x

0

∈Y_η such that

N(L) = span{ϕ1, ϕ2}=: Φ.

Proof. Clearly, ϕ1.ϕ2 are linearly independent. Assumption 2 and Theorem 2.6 imply ϕ1, ϕ2 ∈ Yη. Thus it remains to show Lϕi = 0 for i = 1,2 then the claim follows by Assumption 4. Recall that (v_⋆, v_∞)^⊤ is a stationary solution of (0.22), i.e.

0 =

Av_⋆,xx+cv_⋆,x+S_ωv_⋆ +f(v_⋆) Sωv_∞+f(v_∞)

. By applying the group action a(γ) for γ = (θ, τ)∈ G we obtain

0 =

ARθv⋆,xx(· −τ) +cRθv⋆,x(· −τ) +SωRθv⋆(· −τ) +f(Rθv⋆(· −τ)) SωRθv_∞+f(Rθv_∞)

. (3.44) Differentiating (3.44) w.r.t. θ and evaluating at (θ, τ) = 0 yields

0 =

AS1v⋆,xx+cS1v⋆,x+SωS1v⋆+Df(v⋆)S1v⋆

SωS1v_∞+Df(v_∞)S1v_∞

=L

S1v⋆

S1v_∞

=Lϕ1. Further, differentiating (3.44) w.r.t. τ and evaluating at (θ, τ) = 0 leads to

0 =

−Av⋆,xxx−cv⋆,xx−Sωv⋆,x−Df(v⋆)v⋆,x

0

=−L v⋆,x

0

=Lϕ2.

By Assumption 4 the half-plane {Res > γ} for some γ < 0 does not contain any eigenvalues of L expect for the eigenvalue s = 0. Thus we can assume that the sector Sε,γ from Theorem 3.16 lies in the resolvent set except for the zero eigenvalue.

3.3. THE LINEARIZED OPERATOR L 83 Corollary 3.19. Let the Assumptions 1-4 be satisfied and let 0 < µ < min(µess, µ⋆,2) with µess from Assumption 3 and µ⋆ from Theorem 2.6. Then there are ε > 0, γ < 0 such that

Sε,γ ⊂ρ(L)∪ {0}.

Proof. The claim is a direct consequence of Theorem 3.16 and Assumption 4 by taking ε and |γ|sufficiently small.

As Corollary 3.19 shows the spectrum of L is completely included in the strict left half-plane except the zero eigenvalue. However, since it is an isolated eigenvalue of finite multiplicity we are able to block this neutral direction using the projector onto N(L). This will lead to time decaying estimates of the semigroup on the corresponding orthogonal complement, cf. [32, Thm. 1.5.3.]. For this purpose, we have to take the adjoint ofL into account which will be considered in the following. SinceXη is a Hilbert space we may identify its dual space X_η^∗ with Xη via the Riesz isomorphism. We define the (abstract) adjoint operator of L, cf. [61, Definition IV.11], by

L^∗ :D(L^∗)⊂Xη →Xη, u7→ L^∗u.

For a detailed construction and properties of the adjoint operator L^∗ we refer to [61, IV.11]. Since L has a closed range we have, cf. [61, (11-7)],

R(L)^⊥=N(L^∗), R(L) =N(L^∗)^⊥. (3.45) Lemma 3.20. Let the Assumptions 1-4 be satisfied and 0 < µ < min(µess, µ⋆,2) with µess from Assumption 3 andµ⋆ from Theorem 2.6. Then there are adjoint eigenfunctions ψ1, ψ2 ∈ D(L^∗) such that

i) N(L^∗) = span{ψ1, ψ2}=: Ψ, ii) (ψi, ϕj)Xη =δij, i, j = 1,2, iii) X_η = Φ⊕Ψ^⊥,

iv) there is a continuous projectionP :Xη →Xη ontoΦ, i.e.

P(Φ) = Φ, P(Ψ^⊥) ={0}, P² =P, which is given by

P v :=

X2

i=1

(ψi, v)Xηϕj.

v) the subspace Ψ^⊥ ⊂Xη is invariant under L, i.e. L(Ψ^⊥∩Yη)⊂Ψ^⊥.

Proof. i), ii). Let(·,·) = (·,·)Xη. Lis a Fredholm operator of index 0. Thus by Lemma A.5 the adjoint L^∗ : D(L^∗) → Xη is Fredholm operator of index 0. This implies by Assumption 4 dimN(L^∗) = dimN(L) = 2. Then choose linearly independent ψ^′₁, ψ₂^′ ∈ D(L^∗)such that

N(L^∗) = span{ψ^′₁, ψ₂^′}.

Now by Lemma A.2, the operator Lⁿ is also a Fredholm operator of index 0 on Xη for all n ∈ N. Thus (Lⁿ)^∗ = (L^∗)ⁿ is Fredholm of index 0, which implies by Assumption 4 and Lemma 3.18 that dimN((L^∗)ⁿ) = dimN(Lⁿ) = 2 for n ≥ 2. Hence, L^∗ has no generalized eigenfunctions and therefore ψ^′₁, ψ₂^′ ∈ R/ (L^∗). Now consider the matrix

A :=

(ψ₁^′, ϕ1) (ψ₁^′, ϕ2) (ψ₂^′, ϕ1) (ψ₂^′, ϕ2)

.

We show that A is invertible. Assume the contrary. Then there is z = (z₁, z₂)^⊤ ∈ R² with z^⊤A= 0. This implies

(z1ψ₁^′ +z2ψ₂^′, ϕi) = 0, i= 1,2

and therefore z1ψ^′₁ +z2ψ^′₂ ∈ N(L)^⊥ = R(L^∗) due to (3.45). Then we find w ∈ Xη

such that L^∗w = z1ψ₁^′ +z2ψ₂^′ ∈ N(L^∗). A contradiction since L^∗ has no generalized eigenfunction. Hence A is invertible. Now define

ψ₁ =b₁ψ₁^′ +b₂ψ^′₂, b =A⁻¹ 1

0

, ψ2 =c1ψ₁^′ +c2ψ₂^′, c=A⁻¹

0 1

.

Then ψ1, ψ2 are linearly independent withN(L^∗) = span{ψ1, ψ2} and (ψi, ϕj) =δij. iii). We may write u∈Xη as

u=u− X2

i=1

(ψ_i, u)ϕ_i+ X2

i=1

(ψ_i, u)ϕ_i. Then we have P2

i=1(ψi, u)ϕi ∈ N(L) = Φ and due to ii) ψj, u−

X2

i=1

(ψi, u)ϕi

!

= (ψj, u)− X2

i=1

(ψi, u) (ψj, ϕi) = 0

3.4. THE SEMIGROUP E^TL 85

Im Dokument Stability of Traveling Oscillating Fronts in Parabolic Evolution Equations (Seite 72-85)