Asymptotic study of supercritical surface Quasi-Geostrophic equation in critical space
aa r X i v : . [ m a t h . A P ] F e b Asymptotic study of supercritical surfaceQuasi-Geostrophic equation in critical space
Jamel BENAMEUR and Chaala KATAR ∗ February 23, 2021
Contents H − α
105 Long time decay of global solution 10Abstract
In this paper we prove, if θ ∈ C ([0 , ∞ ) , H − α ( R )) is a globalsolution of supercritical surface Quasi-Geostrophic equation with small initialdata, then k θ ( t ) k H − α decays to zero as time goes to infinity. Fourier analysisand standard techniques are used. The two-dimensional surface quasi-geostrophic equation with supercritical dis-sipation ( QG ) is( QG ) (cid:26) ∂ t θ + | D | α θ + u θ . ∇ θ = 0 θ (0 , x ) = θ ( x ) . where 0 < α < / θ represents the potentialtemperature and u = ( ∂ | D | − , ∂ | D | − ) θ is the fluid velocity. This equation ∗ Department of Mathematics, Faculty of Science of Gab`es, Research LaboratoryMathematics and Applications LR17ES11; Tunisia, [email protected], [email protected]
1s a two-dimensional model of the 3D incompressible Euler equations. In thesuper critical case, there are only partial results. We have global existenceresults when the initial data is small or local existence when the initial data arelarge. The study of global solution is studied by serval researchers, See[6], [8],[10], [12], ...The equation is invariant under the following scaling: θ λ ( t, x ) = λ α − θ ( λ α t, λx )with initial data θ λ = λ α − θ ( λx ), so ˙ H − α is a critical space and we have k λ α − f ( λ. ) k ˙ H − α = k f k ˙ H − α , ∀ λ > Theorem 1.1. ([9]) Let θ ∈ H − α ( R ) . Then, there is a unique time T ∗ ∈ (0 , ∞ ] and a unique solution θ ∈ C ([0 , T ∗ ) , H − α ( R )) of ( QG ) . Our work is devoted to study the maximal solution of ( QG ) system given byTheorem 1.1. First, we study the case of a non-regular solution, where we showthat we have a blow up at the first time of irregularity. Secondly, the case of aregular solution, we show that if moreover the initial condition is small enoughthen the solution decreases towards 0 at infinity. The proof of explosion resultis classic and similar to that of the incompressible Navier-Stokes equations,the only thing to point out that one needs to specify some estimate and thedependence of the associated constants of the index of parametric regularity( C = σ σ C ( α )). But the novelty are the tools used to show the result on theregular solutions where one has to go down in the index of regularity of type˙ H − σ for the high frequencies. Our first main result reads as follows: Theorem 1.2.
Let θ ∈ C ([0 , T ∗ ) , H − α ( R )) be a maximal solution of ( QG ) .If T ∗ < ∞ , then Z T ∗ k| D | α θ ( τ ) k H − α dτ = ∞ . Theorem 1.3.
There is a positif real number ε such that: If θ ∈ H − α ( R ) verifies k θ k H − α < ε , then there exists a global solution of ( QG ) such that θ ∈ C ([0 , ∞ ) , H − α ( R )) ∩ L ( R + , ˙ H − α ( R )) . In addition the following estimate holds k θ k H − α + Z t k| D | α θ k H − α ≤ k θ k H − α . (1) Theorem 1.4. If θ ∈ C ([0 , ∞ ) , H − α ( R )) is a global solution of ( QG ) suchthat k θ (0) k H − α < ε , then lim t →∞ k θ ( t ) k H − α = 0 . emarks . The proof of Theorem 1.4 is done in two steps. In the first, weprove k θ ( t ) k L → t → ∞ , where we use the L -energy estimate k θ ( t ) k L + 2 Z t k| D | α θ ( z ) k L dz ≤ k θ k L and the uniqueness given by Theorem 1.1. In the second step, we prove k θ ( t ) k ˙ H − α → t → ∞ , where we use the uniqueness in H − α ( R ) and θ ∈ L ∞ ([0 , ∞ ) , L ( R )) ∩ L ([0 , ∞ ) , ˙ H − α ( R )) . The rest of this paper is organized as follows: In Section 2, we recall somenotations and we give some important preliminary results. In section 3, weprove a blow-up criterion to the non regular solution of ( QG ) in critical space.Section 4 is devoted to give a proof of Theorem 1.2. In section 5, we prove thatthe norm of global solution in H − α goes to zero at infinity, which is made intwo steps. In this short section, we collect some notations and definitions that will be usedlater. • If A is a subset of R , A denotes the characteristic function of A . • The Fourier transformation is normalized as F ( f )( ξ ) = b f ( ξ ) = Z R exp( − ix.ξ ) f ( x ) dx, ξ = ( ξ , ξ ) ∈ R . • The inverse Fourier formula is F − ( g )( x ) = (2 π ) − Z R exp( iξ.x ) g ( ξ ) dξ, x = ( x , x ) ∈ R . • The convolution product of a suitable pair of function f and g on R isgiven by ( f ∗ g )( x ) := Z R f ( y ) g ( x − y ) dy. • If f : R → R is a function and v = ( v , v , v ) is a vector fields, we setdiv ( f v ) := v. ( ∇ f ) + f div ( v ) . Moreover, if div v = 0 we obtaindiv ( f v ) := v. ( ∇ f ) . If f ∈ S ′ ( R ) such that b f ∈ L loc ( R ), we define | D | σ f by \ | D | α f ( ξ ) = | ξ | α b f ( ξ ) . • For s ∈ R , H s ( R ) denotes the usual non-homogeneous Sobolev space on R and h ., . i H s denotes the usual scalar product on H s ( R ). • For s ∈ R , ˙ H s ( R ) denotes the usual homogeneous Sobolev space on R and h ., . i ˙ H s denotes the usual scalar product on ˙ H s ( R ). • For s >
0, we have (1 + | ξ | ) s ∼ | ξ | s , then k f k H s ∼ k f k L + k| D | s f k L . We can therefore use the equivalent norm q k f k L + k| D | s f k L and the associated scalar product. This scalar product helps us use theproperty h v. ∇ f /f i L = 0 if div ( v ) = 0 , and h v. ∇ f /f i H s = h v. ∇ f /f i ˙ H s . In this section we recall some classical results and we present new technicallemmas.
Proposition 2.1. ([5]) Let H be Hilbert space and ( x n ) be a bounded sequenceof elements in H such that x n → x weakly in H and lim sup n →∞ k x n k ≤ k x k , then lim n →∞ k x n − x k = 0 . Lemma 2.2. ([7]) Let s , s be two real numbers such that s < and s + s > . There exists a constant C = C ( s , s ) , , such that for all f, g ∈ ˙ H s ( R ) ∩ ˙ H s ( R ) , f.g ∈ ˙ H s + s − ( R ) and k f g k ˙ H s s − ≤ C ( k f k ˙ H s k g k ˙ H s + k f k ˙ H s k g k ˙ H s ) . (2) If in addition s < , there exists a constant C = C ( s , s ) such that for all f ∈ ˙ H s ( R ) and g ∈ ˙ H s ( R ) , then f.g ∈ ˙ H s + s − ( R ) and k f g k ˙ H s s − ≤ C k f k ˙ H s k g k ˙ H s . (3)4 emarks . Let α ∈ (0 , / C ( α ) such that:If f ∈ ˙ H ( R ) = L ( R ) and g ∈ ˙ H − α ( R ), then f g ∈ ˙ H − α ( R ) and k f g k ˙ H − α ≤ C ( α ) k f k L k g k ˙ H − α . Lemma 2.4.
Let < α < / , there is a constant C ( α ) such that for θ ∈ H σ + α ( R ) ∩ H − α ( R ) with σ ≥ , we have |h u θ . ∇ θ, θ i H σ | ≤ σ σ C ( α ) k θ k ˙ H − α k θ k H σ + α . Proof.
Using the fact |h u θ . ∇ θ, θ i L = 0, (div u θ = 0), we get |h u θ . ∇ θ, θ i H σ | ≤ Z ξ Z η || ξ | σ − | η | σ || c u θ ( ξ − η ) || c ∇ θ ( η ) | dη | ξ | σ | b θ ( ξ ) | dξ. By using the elementary inequality || ξ | σ − | η | σ | ≤ σ σ − | ξ − η | ( | η | σ − + | ξ − η | σ − )we get |h u θ . ∇ θ, θ i H σ | ≤ | Z ξ Z η | ( | ξ | σ − | η | σ ) | . | b θ ( ξ − η ) || c ∇ θ ( η ) | dη | ξ | σ | b θ ( ξ ) | dξ |≤ σ σ − Z ξ Z η | ξ − η | σ | b θ ( ξ − η ) | . | η || b θ ( η ) | dη | ξ | σ | b θ ( ξ ) | dξ + σ σ − Z ξ Z η | ξ − η || b θ ( ξ − η ) | . | η | σ | b θ ( η ) | dη | ξ | σ | b θ ( ξ ) | dξ ≤ σ σ − (cid:16) Z ξ | ξ | − α ( Z η | ξ − η | σ | b θ ( ξ − η ) || c ∇ θ ( η ) | dη ) dξ (cid:17) / k θ k ˙ H s + α + σ σ − (cid:16) Z ξ | ξ | − α ( Z η | ξ − η || b θ ( ξ − η ) || η | σ | b θ ( η ) | dη ) dξ (cid:17) / k θ k ˙ H σ + α ≤ σ σ k f g k ˙ H − α k θ k ˙ H σ + α , with ( b f = | ξ | . | b θ ( ξ ) | b g = | ξ | σ | b θ ( ξ ) | . Using the product laws, with s + s = 1 − α > (cid:26) s = 1 − α < s = α < Lemma 2.5.
Let < α < / and for θ, ω ∈ H − α ( R ) we have |h u ω . ∇ θ, θ i H − α | ≤ C ( α ) k ω k ˙ H − α k θ k ˙ H − α k θ k ˙ H − α , (4) |h u ω . ∇ θ, θ i H − α | ≤ C ( α ) k ω k ˙ H − α k θ k H − α . (5)5 roof. Proof of equation (4): Put δ = 2 − α , we have |h u ω . ∇ θ, θ i H δ | = |h u ω . ∇ θ, θ i ˙ H δ | = |h| D | δ ( u ω . ∇ θ ) , | D | δ θ i L | = |h| D | δ ( u ω . ∇ θ ) , | D | δ θ i L − h u ω ∇ . | D | δ θ, | D | δ θ i L |≤ C (cid:16) Z ξ | ξ | − α ( Z η | ξ − η | δ | b ω ( ξ − η ) || c ∇ θ ( η ) | dη ) dξ (cid:17) / k θ k ˙ H δ + α + C (cid:16) Z ξ | ξ | − α ( Z η | ξ − η || b ω ( ξ − η ) || η | δ | b θ ( η ) | dη ) dξ (cid:17) / k θ k ˙ H δ + α ≤ C ( k f g k ˙ H − α + k f g k ˙ H − α ) k θ k ˙ H δ + α . with b f = | ξ | δ . | b ω ( ξ ) | b g = | ξ | . | b θ ( ξ ) | b f = | ξ | . | b ω ( ξ ) | b g = | ξ | δ | b θ ( ξ ) | . Using the product laws, withfor ( f , g ) : s + s = 1 − α > s = α < s = 1 − α < f , g ) : s + s = 1 − α > s = 1 − α < s = 0 < . we obtain the desired result.Proof of equation (5): We have |h u ω . ∇ θ, θ i H δ | = |h u ω . ∇ θ, θ i ˙ H δ | = |h| D | δ ( u ω . ∇ θ ) , | D | δ θ i L | = |h| D | δ ( u ω . ∇ θ ) , | D | δ θ i L − h u ω ∇ . | D | δ θ, | D | δ θ i L |≤ C (cid:16) Z ξ | ξ | − α ( Z η | ξ − η | δ | b ω ( ξ − η ) || c ∇ θ ( η ) | dη ) dξ (cid:17) / k θ k ˙ H δ + α + C (cid:16) Z ξ | ξ | − α ( Z η | ξ − η || b ω ( ξ − η ) || η | δ | b θ ( η ) | dη ) dξ (cid:17) / k θ k ˙ H δ + α ≤ C ( k f g k ˙ H − α + k f g k ˙ H − α ) k θ k ˙ H δ + α . with b f = | ξ | δ . | b ω ( ξ ) | b g = | ξ | . | b θ ( ξ ) | b f = | ξ | . | b ω ( ξ ) | b g = | ξ | δ | b θ ( ξ ) | . Using the product laws, withfor ( f , g ) : s + s = 1 − α > s = 0 < s = 1 − α < f , g ) : s + s = 1 − α > s = 1 − α < s = α < , we obtain the desired result. Lemma 2.6.
Let h : [0 , T ] → [0 , ∞ ] be measurable function and σ > , then (cid:16) Z T e − σ ( T − z ) h ( z ) dz (cid:17) ≤ σ − Z T e − σ ( T − z ) h ( z ) dz. Proof.
We begin by denote I = (cid:16) Z T e − σ ( T − z ) h ( z ) dz (cid:17) . We have I = Z T Z T e − σ ( T − z ) e − σ ( T − z ′ ) h ( z ) h ( z ′ ) dzdz ′ . Using the fact h ( z ) h ( z ′ ) ≤ sup( h ( z ) , h ( z ′ ) ) ≤ h ( z ) + h ( z ′ ) , we get I ≤ Z T Z T e − σ ( T − z ) h ( z ) e − σ ( T − z ′ ) dzdz ′ + Z T Z T e − σ ( T − z ) h ( z ′ ) e − σ ( T − z ′ ) dzdz ′ ≤ ( Z T e − σ ( T − z ′ ) dz ′ ) . Z T e − σ ( T − z ) h ( z ) dz + ( Z T e − σ ( T − z ) dz ) Z T h ( z ′ ) e − σ ( T − z ′ ) dz ′ ≤ ( 1 − e − σT σ ) Z T e − σ ( T − z ) h ( z ) dz + ( 1 − e − σT σ ) Z T e − σ ( T − z ) h ( z ) dz ≤ σ Z T e − σ ( T − z ) h ( z ) dz. In this section we prove Theorem 1.2. This proof is done in three steps. Let θ ∈ C ([0 , T ∗ ) , H − α ( R )) be a maximal solution of the system ( QG ). Wesuppose that T ∗ < ∞ and Z T ∗ k| D | α θ k H − α dτ < + ∞ . (6) Step 1:
We prove that θ is bounded in H − α ( R ). For this select a time T ∈ (0 , T ∗ ) such that Z T ∗ T k| D | α θ k H − α dτ < C ( α ) . Then, by using (4) we7et: for all t ∈ [ T , T ∗ ) k θ ( t ) k H − α + 2 Z tT k θ k H − α dτ ≤ k θ ( T ) k H − α + 2 Z tT | < u θ ( τ ) . ∇ θ ( τ ) /θ ( τ ) > ˙ H − α | dτ ≤ k θ ( T ) k H − α + 2 C ( α ) Z tT k θ ( τ ) k ˙ H − α k θ k H − α dτ ≤ k θ ( T ) k H − α + 2 C ( α ) sup z ∈ [ T ,t ] k θ ( z ) k ˙ H − α Z tT k θ k H − α dτ ≤ k θ ( T ) k H − α + 12 sup z ∈ [ T ,t ] k θ ( z ) k ˙ H − α . Then k θ ( t ) k H − α ≤ k θ ( T ) k H − α , ∀ t ∈ [ T , T ∗ )Put M = max( sup z ∈ [0 ,T ] k θ ( z ) k H − α ; √ k θ ( T ) k H − α ) . Step 2:
We want to prove θ is a Cauchy family at T ∗ in L ( R ). For t < t ′ ∈ [ T , T ∗ ), we have θ ( t ) − θ ( t ′ ) = − Z t ′ t | D | α θ − Z t ′ t u θ ∇ θ. Then k θ ( t ) − θ ( t ′ ) k L ≤ Z t ′ t k| D | α θ − u θ ∇ θ k L . By using Lemma 2.2 with s = 1 − α and s = 2 α , we get k θ ( t ) − θ ( t ′ ) k L ≤ Z t ′ t k θ k H α + Z t ′ t k u θ ∇ θ k L ≤ Z t ′ t k θ k ˙ H α + C ( α ) Z t ′ t k θ k ˙ H α k θ k ˙ H − α ≤ Z t ′ t k θ k ˙ H α + C ( α ) M Z t ′ t k θ k ˙ H α ≤ (1 + C ( α ) M ) Z t ′ t k θ k ˙ H α ≤ (1 + C ( α ) M ) Z t ′ t k θ k H α ≤ (1 + C ( α ) M ) Z t ′ t k θ k H − α ≤ (1 + C ( α ) M ) M ( t ′ − t ) . Then θ ( t ) is a Cauchy type at T ∗ . As L is Banach space, then there is anelement θ ∗ in L such that θ ( t ) → θ ∗ in L if t goes to T ∗ . As θ is bounded inthe Banach space H − α ( R ), then θ ∗ ∈ H − α ( R ).8 tep 3: In this step we prove that lim t → T ∗ k θ ( t ) − θ ∗ k H − α = 0.By step 2 and by interpolation, since θ ( t ) is bounded in H − α we havelim t → T ∗ k θ ( t ) − θ ∗ k H s = 0 , ∀ ≤ s < − α. (7)Let 0 < t < r < T ∗ and a positive real sequence ( s k ) k such that 1 < s k < − α and s k ր − α . We have k θ ( r ) k H sk + 2 Z rt k| D | α θ k H sk = k θ ( t ) k H sk − Z rt h u θ . ∇ θ, θ i H sk | {z } I . By using Lemma 2.4, with σ = s k , we get | I | ≤ s k s k C ( α ) Z rt k θ k ˙ H − α k θ k H sk + α ≤ (2 − α )2 − α C ( α ) Z rt M k| D | α θ k H sk ≤ (2 − α )2 − α C ( α ) M Z rt k| D | α θ k H sk ≤ (2 − α )2 − α C ( α ) M Z rt k| D | α θ k H − α . Then k θ ( t ) k H sk ≤ k θ ( r ) k H sk + (cid:16) − α )2 − α C ( α ) M + 2 (cid:17) Z rt k| D | α θ k H − α . We pass r to T ∗ , taking into account the equation (7), we obtain k θ ( t ) k H sk ≤ k θ ∗ k H sk + (cid:16) − α )2 − α C ( α ) M + 2 (cid:17) Z T ∗ t k| D | α θ k H − α ≤ k θ ∗ k H − α + (cid:16) − α )2 − α C ( α ) M + 2 (cid:17) Z T ∗ t k| D | α θ k H − α . By Monotone convergence theorem, if k → ∞ , we get k θ ( t ) k H − α ≤ k θ ∗ k H − α + (cid:16) − α )2 − α C ( α ) M + 2 (cid:17) Z T ∗ t k| D | α θ k H − α . Then, we pass lim sup t T ∗ , taking into account the equation (6), we obtainlim sup t T ∗ k θ ( t ) k H − α ≤ k θ ∗ k H − α . So, by Proposition 2.1 we get lim t → T ∗ k θ ( t ) − θ ∗ k H − α = 0. Therefore the systemof quasi geostrophic starting by θ ∗ has a unique solution extends θ , which isabsurd. 9 Global solution in H − α In this section we prove Theorem 1.3. Precisely, we prove that if k θ k ˙ H − α < C , we get a global solution in C ( R + , H − α ( R )) satisfying (1). Using Lemma2.4 to estimate the non linear part of ( QG ), we obtain: k θ ( t ) k H − α + 2 Z t k| D | α θ k H − α ≤ k θ k H − α + 2 C Z t k θ k H − α k| D | α θ k H − α . Now, consider the time T defined as follows: T = sup ≤ t ≤ T ∗ { sup ≤ z ≤ t k θ ( z ) k H − α < k θ k H − α } . By continuity of the function ( t → k θ ( t ) k H − α ) we obtain T > ≤ t < T , we have k θ ( t ) k H − α + Z t k| D | α θ k H − α dτ ≤ k θ k H − α . (8)Again by continuity of the function ( t → k θ ( t ) k H − α ) and the above inequalitywe get T = T ∗ and Z T ∗ k| D | α θ k H − α dτ < ∞ . Hence T ∗ = ∞ , and (1) is given by (8), which complete the proof of Theorem1.3. In this section, we prove Theorem 1.4. This proof is done in two steps.
Step 1 : In this step we shall prove that lim t →∞ k θ ( t ) k L = 0.We beginning by by recalling the L energy estimate: k θ k L + 2 Z t k| D | α θ k L ≤ k θ k L . (9)For δ >
0, we define the following functions w δ = A δ ( D ) θ = F − ( B (0 ,δ ) b θ ) : the low frequency part of θ,v δ = B δ ( D ) θ = F − ((1 − B (0 ,δ ) ) b θ ) : the high frequency part of θ. and w δ = A δ ( D ) θ ,v δ = B δ ( D ) θ . We apply the operator A δ ( D ) to the first equation of system ( QG ), we obtain ∂ t w δ + | D | α w δ + A δ ( D )( u θ . ∇ θ ) = 0 . w δ in last equation and integrating with respectto time, we obtain k w δ ( t ) k L + 2 Z t k| D | α w δ k L ≤ k w δ k L + 2 Z t |h A δ ( D )( u θ ∇ θ ) , w δ i L | . Using the fact A δ ( D ) = A δ ( D ) and h A δ ( D ) f, g i L = h f, A δ ( D ) g i L , we get |h A δ ( D )( u θ ∇ θ ) , w δ i L | = |h u θ ∇ θ, w δ i L |≤ Z | ξ | <δ |F (div( θu θ ))( ξ ) | . | b w δ ( ξ ) dξ |≤ Z | ξ | <δ | ξ | . |F ( θu θ )( ξ ) | . | b w δ ( ξ ) dξ |≤ Z | ξ | <δ | ξ | − α . | ξ | α − . |F ( θu θ )( ξ ) | . | b w δ ( ξ ) dξ |≤ δ − α Z | ξ | <δ | ξ | α − . |F ( θu θ )( ξ ) | . | b w δ ( ξ ) dξ |≤ δ − α (cid:16) Z | ξ | α − . |F ( θu θ )( ξ ) | dξ (cid:17) / k b w δ k L ≤ δ − α k θu θ k ˙ H α − k b w δ k L . Applying the product law in homogeneous Sobolev with s = s = α and usingthe fact | c u θ ( ξ ) | = | b θ ( ξ ) | , we get k θu θ k ˙ H α − ≤ C ( α ) k θ k H α and |h A δ ( D )( u θ ∇ θ ) , w δ i L | ≤ C ( α ) δ − α k θ k H α k w δ k L . Using k A δ ( D ) f k L ≤ k f k L and the L energy estimate (9), we get k w δ k L ≤k θ k L and k w δ ( t ) k L + 2 Z t k| D | α w δ k L ≤ k w δ k L + 2 C ( α ) δ − α k θ k L Z t k θ k H α ≤ k w δ k L + C ( α ) δ − α k θ k L := ε δ . Clearly lim δ → + ε δ = 0, thenlim δ → + sup t ≥ k w δ k L = 0 and lim δ → + Z ∞ k w δ k H α = 0 . Let ε > ε < / C ), then there is δ > < δ ≤ δ , we havesup t ≥ k w δ k L < ε/ Z ∞ k w δ k H α < ε/ . (10) In the following we fix a real δ ∈ (0 , δ ].On the other hand v δ verifies ∂ t v δ + | D | α v δ + B δ ( D )( u θ . ∇ θ ) = 0 , v δ = e − t | D | α v δ − Z t e − ( t − z ) | D | α B δ ( D )( u θ . ∇ θ ) dz. Taking the norm in ˙ H − σ ( R ), with σ = 2 − α >
0, we get k v δ ( t ) k ˙ H − σ ≤ k e − t | D | α v δ k ˙ H − σ + k Z t e − ( t − z ) | D | α B δ ( D )( u θ . ∇ θ ) dz k ˙ H − σ ≤ δ − σ e − tδ α k b v δ k L + (cid:16) Z | ξ | >δ | ξ | − σ ( Z t e − ( t − z ) | ξ | α |F ((div( θu θ )))( z, ξ ) | dz ) dξ (cid:17) / ≤ δ − σ e − tδ α k b θ k L + (cid:16) Z | ξ | >δ | ξ | − σ ( Z t e − ( t − z ) | ξ | α |F (( θu θ ))( z, ξ ) | dz ) dξ (cid:17) / . Using Lemma 2.6, we get k v δ ( t ) k ˙ H − σ ≤ δ − σ e − tδ α k b θ k L + (cid:16) Z | ξ | >δ | ξ | − σ − α Z t e − ( t − z ) | ξ | α |F (( θu θ ))( z, ξ ) | dzdξ (cid:17) / . Using the fact 2 − σ − α = 2(2 α −
1) and the product law in homogeneousSobolev space with s = s = α , we get k v δ ( t ) k ˙ H − σ ≤ δ − σ e − tδ α k b θ k L + (cid:16) Z t e − ( t − z ) δ α Z | ξ | >δ | ξ | α − |F (( θu θ ))( z, ξ ) | dξdz (cid:17) / ≤ δ − σ e − tδ α k b θ k L + (cid:16) Z t e − ( t − z ) δ α k ( θu θ )( z ) k H α − dz (cid:17) / ≤ δ − σ e − tδ α k b θ k L + C ( α ) (cid:16) Z t e − ( t − z ) δ α k θ ( z ) k H α dz (cid:17) / Using the fact Z ∞ (cid:16) δ − σ e − tδ α k b θ k L (cid:17) dt = δ − σ δ − α k b θ k L Z ∞ h(cid:16) Z t e − ( t − z ) δ α k θ ( z ) k H α dz (cid:17) / i dt ≤ (cid:16) Z t e − tδ α dt (cid:17)(cid:16) Z ∞ k θ ( t ) k H α dt (cid:17) ≤ δ − α k b θ k L we obtain v δ ∈ L ( R + , ˙ H − σ ( R )) and Z ∞ k v δ ( t ) k H − σ dt ≤ (cid:16) δ − σ δ − α k b θ k L δ − α k b θ k L (cid:17) := M δ . The L energy estimate (9) and inequality (10) give v δ = θ − w δ ∈ L ( R + , ˙ H α ( R )).Therefore v δ ∈ L ( R + , ˙ H − σ ( R ) ∩ ˙ H α ( R )) and by the injection˙ H − σ ( R ) ∩ ˙ H α ( R ) ֒ → ˙ H ( R ) = L ( R ) ,
12e get v δ ∈ L ( R + , L ( R )). Now, we define the set of times E δ = { t ≥ , k v δ ( t ) k L > ε } . Then ( ε λ ( E δ ) ≤ Z E δ k v δ ( t ) k L dt ≤ Z ∞ k v δ ( t ) k L dt < ∞ . Then, we can define the finite time T ε = ( 2 ε ) Z ∞ k v δ ( t ) k L dt and λ ( E δ ) ≤ T ε . For r >
0, there exists t ∈ [0 , T ε + r ] such that t / ∈ E δ and it results that k v δ ( t ) k L ≤ ε . (11)The fact θ = v δ + w δ and equations (10), (11) give that k θ ( t ) k L ≤ ε. Using the uniqueness of solution ( γ ( t ) = θ ( t + t )) in H − α for the followingsystem ( QG t ) (cid:26) ∂ t γ + | D | α γ + u γ . ∇ γ = 0 γ (0 , x ) = θ ( t , x ) . we get k θ ( t + t ) k L + 2 Z t k| D | α θ ( t + z ) k L dz ≤ k θ ( t ) k L < ε, ∀ t ≥ . Then k θ ( t ) k L → t → ∞ , which complete the first step. Step 2:
We shall prove that lim t → k θ ( t ) k ˙ H − α = 0 . As k θ k H − α < C , then Theorem 1.3 gives θ ∈ L ∞ ( R + , H − α ( R )) ∩ L ( R + , ˙ H − α ( R )) . Using the interpolation inequality k θ ( t ) k ˙ H − α ≤ k θ ( t ) k α − α L k θ ( t ) k − α − α ˙ H − α , we get θ ∈ L − α − α ( R + , ˙ H − α ( R )) . Let ε be fixed positif real number and define the set of times F ε = { t ≥ t , k θ ( t ) k ˙ H − α ≥ ε } , t is the time given by the first step. Then ε − α − α λ ( F ε ) ≤ Z F ε k θ ( t ) k − α − α ˙ H − α dt ≤ Z ∞ t k θ ( t ) k − α − α ˙ H − α dt < ∞ . Then, we can define the finite time t ε = ε − − α − α Z ∞ k θ ( t ) k − α − α ˙ H − α dt. We have λ ( F ε ) ≤ t ε . Then, there is a time t ∈ [ t , t + t ε + 1] such that t / ∈ F ε and it results that k θ ( t ) k ˙ H − α ≤ ε. As t ≥ t , then k θ ( t ) k L ≤ ε and k θ ( t ) k H − α ≤ ε. By choosing ε < / C , we get k θ ( t ) k H − α + Z tt k| D | α θ ( z ) k H − α dz ≤ ε , ∀ t ≥ t . which complete the second step, and the proof of Theorem 1.4 is finished. References [1] J. Benameur,
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