Asymptotic study of a global solution of super-critical Quasi-Geostrophic equation
aa r X i v : . [ m a t h . A P ] F e b ASYMPTOTIC STUDY OF A GLOBAL SOLUTION OFSUPER-CRITICAL QUASI-GEOSTROPHIC EQUATION
CHAALA KATAR
Abstract.
In this paper, we study the super-critical Quasi-Geostrophic equation inGevrey-Sobolev space. We prove the local existence of ( QG ) for any large initial dataand we give an exponential type of Blow-up to the solution. Moreover, we establish theexistence global for a small initial data and we show that k θ k H sa,α − decays to zero astime goes to infinity. Fourier analysis and standard techniques are used. Contents
1. Introduction 12. Notations and preliminaries results 32.1. Notations 32.2. Preliminaries results 32.3. Local well-posedness 82.4. Exponential type explosion 143. Global Solution 183.1. Long time decay of global solution 204. Appendix 22References 241.
Introduction
We are concerned with the following two-dimensional quasi-geostrophic equation ( QG ):( QG ) ∂ t θ + κ | D | α θ + u θ . ∇ θ = 0 u θ = ( u θ , u θ ) = ( − ∂ | D | − θ, ∂ | D | − θ ) θ (0 , x ) = θ ( x ) . Here 0 < α < / κ > θ represents the potential temperature and u = ( ∂ | D | − , ∂ | D | − ) θ is the fluid velocity.Our aim in this paper is to study of explosions for non-smooth solutions of ( QG ) in finitemaximal time. Remark 1.1.
The linear system of ( QG ) is ( LQG ) ( ∂ t θ + | D | α θ = 0 θ (0 , x ) = θ ( x ) ∈ H s . Since θ ∈ H s , the solution of the ( LQG ) is θ = e − t | D | α θ ∈ H st, (2 α ) − .with H sa,σ is Gevrey Sobolev space which is defined as follow: for a > , σ > and s > , H sa,σ ( R d ) = { u ∈ L ( R d ) / (1 + | ξ | ) s/ e a | ξ | /σ ∈ L ( R d ) } . equipped by the norm k u k H sa,σ = (cid:16) Z R d (1 + | ξ | ) s | b u ( ξ ) | e a | ξ | /σ dξ (cid:17) and the associated inner product h f, g i ˙ H sa,σ = h e a | D | /σ f, e a | D | /σ g i ˙ H s The explosion of type exponential has been studied in the previous work of Benameur[1]. The author use the Sobolev-Gevrey space to get better explosion result.More precisely, we use the same space in order to prove that the type of explosion is dueto the chosen space not to nonlinear part of ( QG ).On the other wise, we study the asymptotic behavior of the two-dimensional quasi-geostrophic equations with super critical dissipation. In literature, the global regularityhas been shown when the initial data is small in spaces B − α , [4], H s , s > B s , ∞ with s > − α [12]..We finish by explaining why we choose σ = α − in the definition of Sobolev Gevrey space,the reason appears in the non-linear estimate of lemma 2.4. In the proof of this lemma wehave seen that in the case σ = α − there is a perfect balance between the nonlinear termand the dissipation when we have a L ( H s + αa,α − ) control.Let us fix k = 1 for the rest of the paper.The following are our main theorems. Theorem 1.2.
Let a, s, α ∈ R such that a > , s > and < α < . Let θ ∈ H sa,α − (cid:0) R (cid:1) . There is a unique time T ∗ ∈ (0 , ∞ ] and a unique solution θ ∈ C ([0 , T ∗ ) , H sa,α − (cid:0) R (cid:1) ) of ( QG ) Moreover, if T ∗ < ∞ , then C ( T ∗ − t ) exp (cid:0) aC ( T ∗ − t ) αs (cid:1) ≤ k θ ( t ) k H sa,α − (1.1) where C and C are positif constants. For the small initial data, the global existence is given by the following theorem.
Theorem 1.3.
Let θ ∈ H sa,α − ( R ) .If k θ k H sa,α − < √ C then there exists a global solution of ( QG ) such that θ ∈ C ([0 , ∞ ) , H sa,α − ( R )) Where C is positif constant. Moreover, we have for all t ≥ k θ k H sa,α − + Z t k| D | α θ k H sa,α − ≤ k θ k H sa,α − . (1.2) Theorem 1.4.
Let s > , a > and < α < .If θ ∈ C ( R + , H sa,α − ( R )) is a global solution of ( QG ) then lim t →∞ k θ ( t ) k H sa,α − = 0 . SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 3
The remaining part of the article is organized as follows. In Section 2, we present somenotations and we show preliminary results which will be very useful for this paper.In section 4, we prove the theorem 1.2 and we have in subsection 2.4 a blow up result oftype exponential. Then in the subsection 3.1, we state that the norm of global solution in H sa,α − ( R ) goes to zero at infinity.2. Notations and preliminaries results
Notations. • F ( f )( ξ ) = b f ( ξ ) = R R exp( − ixξ ) f ( x ) dx ; ξ = ( ξ , ξ ) ∈ R • 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 ). • \ | D | α f ( ξ ) = | ξ | α b f ( ξ ) • For σ ∈ R , the Fourier space is defined by X σ ( R ) = { f ∈ S ′ ( R ) / b f ∈ L loc and | D | σ b f ∈ L ( R ) } . • k f k X σ = R R | ξ | σ | b f ( ξ ) | dξ is a norm on X σ ( R ). • u θ − u θ = u θ − θ . • If ( B, k . k ) be a Banach space: C b ( I, B ) = space of continuous bounded functionsfrom an interval I to B .2.2. Preliminaries results.Proposition 2.1. ( [3] ) Let H be Hilbert space and ( x n ) be a bounded sequence of elementsin 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. ( [1] ) Let s ≥ , a > and σ > . Then, there is a constant C = C ( s ) suchthat for all f, g ∈ H sa,σ ( R ) , we have k f g k H sa,σ ≤ C (cid:16) k e aσ | D | /σ f k X k g k H sa,σ + k f k H sa,σ k e aσ | D | /σ g k X (cid:17) . Moreover, if s > d/ , we have H sa,σ is algebra and k f g k H sa,σ ≤ c ( s ) k f k H sa,σ k g k H sa,σ . C. KATAR
Lemma 2.3. ( [2] ) 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 . Lemma 2.4.
Let s > , a > , α ∈ (0 , and for every θ, ω ∈ H s + αa,α − ( R ) , there is aconstant C = C s,a,α such that (2.1) |h u θ . ∇ ω, ω i H sa,α − | ≤ C (cid:16) k e aα | D | α θ k X k| D | α ω k H sa,α − + k| D | α θ k H sa,α − k e aα | D | α ω k X (cid:17) k ω k H sa,α − , (2.2) |h u θ . ∇ θ, θ i H sa,α − | ≤ C k e aα | D | α θ ) k X k| D | α θ k H sa,α − k θ k H sa,α − , (2.3) |h u θ . ∇ θ, θ i H sa,α − | ≤ C k| D | α θ k H sa,α − k θ k H sa,α − , (2.4) |h u θ . ∇ ω, ω i H sa,α − | ≤ C k θ k H sa,α − k| D | α ω k H sa,α − k ω k H sa,α − . Proof.
Clearly equation (2.2) is a particular case of (2.1). We start by proving the firstequation. • Proof of (2.1): Using the fact h u θ . ∇ ω, ω i L = 0 and h f, g i H sa,σ = h f, g i L + h f, g i ˙ H sa,σ , weget < u θ . ∇ ω, ω > H sa,σ = < u θ . ∇ ω, ω > ˙ H sa,σ . Using again h u θ . ∇| D | s e a | D | α ω, | D | s e a | D | α ω i L = 0 and Cauchy-Schwartz inequality, weobtain |h u θ . ∇ ω, ω i ˙ H sa,α − | ≤ Z ξ Z || ξ | s e a | ξ | α − | η | s e a | η | α | . | b θ ( ξ − η ) || d ∇ ω ( η ) || ξ | s e a | ξ | α | b ω ( − ξ ) | dξ ≤ (cid:16) Z ξ ( Z || ξ | s e a | ξ | α − | η | s e a | η | α | . | b θ ( ξ − η ) || d ∇ ω ( η ) | dη ) dξ (cid:17) / k ω k ˙ H sa,α − . For ξ, η ∈ R , there is z ∈ [min( | ξ | , | η | ) , max( | ξ | , | η | )] such that | ξ | s e a | ξ | α − | η | s e a | η | α = ( | ξ | − | η | ) (cid:16) sz s − + aαz s + α − (cid:17) e az α . Using the fact z ≤ max( | ξ | , | η | ) ≤ | ξ − η | + | η | ≤ | ξ − η | , | η | ) and the fact (see [1]) | ξ | α ≤ max( | ξ − η | , | η | ) α + α min( | ξ − η | , | η | ) α , we get sz s − + aαz s + α − ≤ s + α − ( s + aα ) (cid:16) max( | ξ − η | , | η | ) s − + max( | ξ − η | , | η | ) s + α − (cid:17) e az α ≤ (cid:26) e a | η | α , if | ξ | < | η | e a | ξ | α , if | ξ | > | η |≤ (cid:26) e a | η | α , if | ξ | < | η | e a max( | ξ − η | , | η | ) α + aα min( | ξ − η | , | η | ) α , if | ξ | > | η |≤ e a max( | ξ − η | , | η | ) α + aα min( | ξ − η | , | η | ) α . Then (cid:12)(cid:12)(cid:12) | ξ | s e a | ξ | α −| η | s e a | η | α (cid:12)(cid:12)(cid:12) ≤ C | ξ − η | (cid:16) | ξ − η | s − + | ξ − η | s + α − (cid:17) e a | ξ − η | α e aα | η | α , if | ξ − η | > | η | C | ξ − η | (cid:16) | η | s − + | η | s + α − (cid:17) e a | η | α e aα | ξ − η | α , if | ξ − η | < | η | SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 5 with C = 2 s + α ( s + α ). Therefore |h u θ . ∇ ω, ω i ˙ H sa,α − | ≤ C (cid:16) X k =1 I k (cid:17) k ω k ˙ H sa,α − , with I = (cid:16) Z ξ ( Z η | ξ − η | s e a | ξ − η | α | b θ ( ξ − η ) | e aα | η | α | η | . | b ω ( η ) | dη ) dξ (cid:17) / := k F ∗ G k L F = | ξ | s e a | ξ | α | b θ ( ξ ) | , G = | ξ | e aα | ξ | α | b ω ( ξ ) | I = (cid:16) Z ξ ( Z η | ξ − η | s + α e a | ξ − η | α | b θ ( ξ − η ) | e aα | η | α | η | . | b ω ( η ) | dη ) dξ (cid:17) / := k F ∗ G k L F = | ξ | s + α e a | ξ | α | b θ ( ξ ) | , G = | ξ | e aα | ξ | α | b ω ( ξ ) | I = (cid:16) Z ξ ( Z η | ξ − η | e aα | ξ − η | α | b θ ( ξ − η ) | e a | η | α | η | s | b ω ( η ) | dη ) dξ (cid:17) / := k F ∗ G k L F = | ξ | e aα | ξ | α | b θ ( ξ ) | , G = | ξ | s e a | ξ | α | b ω ( ξ ) | I = (cid:16) Z ξ ( Z η | ξ − η | e aα | ξ − η | α | b θ ( ξ − η ) | . | η | s + α e a | η | α | b ω ( η ) | dη ) dξ (cid:17) / := k F ∗ G k L F = | ξ | e aα | ξ | α | b θ ( ξ ) | , G = | ξ | s + α e a | ξ | α | b ω ( ξ ) | . Young inequality implies |h u θ . ∇ ω, ω i ˙ H sa,σ | ≤ C (cid:16) k F k L k G k L + k F k L k G k L + k F k L k G k L + k F k L k G k L (cid:17) k ω k ˙ H sa,α − . As 0 ≤ α ≤ s , then | ξ | s ≤ (cid:26) | ξ | α if | ξ | < | ξ | s + α if | ξ | > ≤ (1 + | ξ | ) s | ξ | α . By the above inequality we obtain k F k L = k θ k ˙ H sa,α − ≤ k| D | α θ k H sa,α − k G k L = k ω k ˙ H sa,α − ≤ k| D | α ω k H sa,α − . which imply the desired result. • Proof of (2.3): By using equation (2.2), it suffices to prove k e aα | D | α θ ) k X ≤ C k θ ) k H sa,α − .For this, we write k e aα | D | α θ k X = Z R | ξ | e aα | ξ | α | b θ ( ξ ) | dξ = Z R | ξ | e aα | ξ | α (1 + | ξ | ) s/ e a | ξ | α (1 + | ξ | ) s/ e a | ξ | α | b θ ( ξ ) | dξ ≤ (cid:16) Z R | ξ | e aα | ξ | α (1 + | ξ | ) s/ e a | ξ | α dξ (cid:17) / k θ k ˙ H sa,α − , which gives (2.3). • Proof of (2.4): We have |h u θ . ∇ ω, ω i ˙ H sa,σ | ≤ Z ξ Z || ξ | s e a | ξ | α − | η | s e a | η | α | . | b θ ( ξ − η ) || d ∇ ω ( η ) || ξ | s e a | ξ | α | b ω ( − ξ ) | dξ. C. KATAR
By using the same steps of the above proof, we get: For ξ, η ∈ R , (cid:12)(cid:12)(cid:12) | ξ | s e a | ξ | α −| η | s e a | η | α (cid:12)(cid:12)(cid:12) ≤ C | ξ − η | (cid:16) | ξ − η | s − + | ξ − η | s + α − (cid:17) e a | ξ − η | α e aα | η | α , if | ξ − η | > | η | C | ξ − η | (cid:16) | η | s − + | η | s + α − (cid:17) e a | η | α e aα | ξ − η | α , if | ξ − η | < | η | . Using the fact | ξ − η | s + α − ≤ c | ξ − η | s − ( | ξ | α + | η | α ) we get (cid:12)(cid:12)(cid:12) | ξ | s e a | ξ | α −| η | s e a | η | α (cid:12)(cid:12)(cid:12) ≤ C | ξ − η | (cid:16) | ξ − η | s − + | ξ − η | s − ( | ξ | α + | η | α ) (cid:17) e a | ξ − η | α e aα | η | α , if | ξ − η | > | η | C | ξ − η | (cid:16) | η | s − + | η | s + α − (cid:17) e a | η | α e aα | ξ − η | α , if | ξ − η | < | η | . Therefore |h u θ . ∇ ω, ω i ˙ H sa,σ | ≤ C X k =1 J k , with J = Z ξ ( Z η | ξ − η | s e a | ξ − η | α | b θ ( ξ − η ) | e aα | η | α | η | . | b ω ( η ) | dη ) | ξ | s e a | ξ | α | b ω ( − ξ ) | dξJ = Z ξ ( Z η | ξ − η | s e a | ξ − η | α | b θ ( ξ − η ) | e aα | η | α | η | . | b ω ( η ) | dη ) | ξ | s + α e a | ξ | α | b ω ( − ξ ) | dξJ = Z ξ ( Z η | ξ − η | s e a | ξ − η | α | b θ ( ξ − η ) | e aα | η | α | η | α . | b ω ( η ) | dη ) | ξ | s e a | ξ | α | b ω ( − ξ ) | dξJ = Z ξ ( Z η | ξ − η | e aα | ξ − η | α | b θ ( ξ − η ) | e a | η | α | η | s | b ω ( η ) | dη ) | ξ | s e a | ξ | α | b ω ( − ξ ) | dξJ = Z ξ ( Z η | ξ − η | e aα | ξ − η | α | b θ ( ξ − η ) | e a | η | α | η | s + α | b ω ( η ) | dη ) | ξ | s e a | ξ | α | b ω ( − ξ ) | dξ. Combining Cauchy-Schwartz and Young inequalities, we get J ≤ k θ k ˙ H sa,α − k e aα | D | α ω k X k ω k ˙ H sa,α − J ≤ k θ k ˙ H sa,α − k e aα | D | α ω k X k ω k ˙ H s + αa,α − J ≤ k θ k ˙ H sa,α − k e aα | D | α ω k X α k ω k ˙ H sa,α − J ≤ k e aα | D | α θ k X k ω k ˙ H sa,α − k ω k ˙ H sa,α − J ≤ k e aα | D | α θ k X k ω k ˙ H s + αa,α − k ω k ˙ H sa,α − . Using the fact | ξ | s ≤ (1 + | ξ | s ) | ξ | α , (0 < α < s ) and Z R | ξ | e aα | ξ | α (1 + | ξ | ) s e a | ξ | α dξ, Z R | ξ | α e aα | ξ | α (1 + | ξ | ) s e a | ξ | α dξ < ∞ we obtain the result. Lemma 2.5.
Let < α < , there is a constant C ( s ) such that for θ ∈ H s + αa,α − ( R ) with s ≥ , we have We have (2.5) |h u θ . ∇ θ, θ i H sa,α − | ≤ C k θ k H sa,α − k| D | α θ k H sa,α − , proof. Using h u θ . ∇| D | s e a | D | α θ, | D | s e a | D | α θ i L = 0, we obtain |h u θ . ∇ θ, θ i ˙ H sa,α − | ≤ Z ξ Z η || ξ | s e a | ξ | α − | η | s e a | η | α | . | b θ ( ξ − η ) || c ∇ θ ( η ) || ξ | s e a | ξ | α | b α ( − ξ ) | dηdξ. SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 7
The inequality e a | ξ | α ≤ e a | ξ − η | α e a | η | α implies (cid:12)(cid:12)(cid:12) | ξ | s e a | ξ | α −| η | s e a | η | α (cid:12)(cid:12)(cid:12) ≤ C | ξ − η | (cid:16) | ξ − η | s − + | ξ − η | s + α − (cid:17) e a | ξ − η | α e a | η | α , if | ξ − η | > | η | C | ξ − η | (cid:16) | η | s − + | η | s + α − (cid:17) e a | η | α e a | ξ − η | α , if | ξ − η | < | η | Therefore |h u θ . ∇ θ, θ i H sa,α − | ≤ C X k =1 I k , with I ≤ Z ξ ( Z η | ξ − η | s e a | ξ − η | α | b θ ( ξ − η ) | . | η | e a | η | α | b θ ( η ) | dη ) | ξ | s e a | ξ | α | b θ ( ξ ) | dξI ≤ Z ξ ( Z η | ξ − η | s + α e a | ξ − η | α | b θ ( ξ − η ) | . | η | e a | η | α | b θ ( η ) | dη ) | ξ | s | b θ ( ξ ) | dξI ≤ Z ξ ( Z η | ξ − η | e a | ξ − η | α | b θ ( ξ − η ) | . | η | s e a | η | α | b θ ( η ) | dη ) | ξ | s e a | ξ | α | b θ ( ξ ) | dξI ≤ Z ξ ( Z η | ξ − η | e a | ξ − η | α | b θ ( ξ − η ) | . | η | s + α e a | η | α | b θ ( η ) | dη ) | ξ | s | b θ ( ξ ) | dξ. Then |h u θ . ∇ θ, θ i H sa,α − | ≤ C (cid:0) k f g k ˙ H − α + k f g k ˙ H − α (cid:1) k θ k ˙ H s + α with b f = | ξ | e a | ξ | α | b θ ( ξ ) | , b g = | ξ | s e a | ξ | α | b θ ( ξ ) | , b f = | ξ | e a | ξ | α | b θ ( ξ ) | , b g = | ξ | s + α e a | ξ | α | b θ ( ξ ) | . Using the product laws in homogenous Sobolev space for f , g , with s + s = 1 − α > ( s = 1 − α < s = α < f , g , with s + s = 1 − α > ( s = 1 − α < s = 0 < |h u θ . ∇ θ, θ i H sa,α − | ≤ C (cid:0) k f k ˙ H − α k g k ˙ H α + k f k ˙ H − α k g k ˙ H (cid:1) k θ k ˙ H s + α ≤ (cid:0) k θ k ˙ H − αa,α − k θ k ˙ H s + αa,α − + k θ k ˙ H − αa,α − k θ k ˙ H s + αa,α − (cid:1) k θ k ˙ H s + αa,α − . ≤ (cid:0) k θ k H − αa,α − k θ k H s + αa,α − + k θ k H − αa,α − k θ k H s + αa,α − (cid:1) k θ k H s + αa,α − . Now, using the fact 2 − α ≤ < s and 2 − α ≤ < s to obtain the desired result. Remark 2.6.
For a > , s > and σ > , we have e − a | D | /σ S ( R ) = H sa,σ ( R ) . C. KATAR
Indeed, it suffices to write S ( R ) = H s ( R ) ∀ s > . Local well-posedness.
This subsection is devoted to the proof of the local well-posedness of the system ( QG ), which we give the proof the following theorem. Theorem 2.7.
Let a, s, α ∈ R such that a > , s > and < α < . Let θ ∈ H sa,α − (cid:0) R (cid:1) . There is a time
T > and a unique solution θ ∈ C (cid:16) [0 , T ] , H sa,α − (cid:0) R (cid:1)(cid:17) of ( QG ) . Proof.
Proving this theorem requires four steps:- First, we apply the method fixed point theorem so as to solve a approximating system( QG ) k of ( QG ).- Second, we prove that for uniform T sufficiently small, the sequence ( θ k ) is bounded in C (cid:16) [0 , T ]; H sa,α − (cid:17) .- Third, we establish that for T sufficiently small, ( θ k ) is a Cauchy sequence in L ∞ ([0 , T ]; H s − a,α − ).- Finally, we check that the limit of this sequence is a solution of ( QG ) and that it belongsto C ([0 , T ]; H sa,α − ). Step1: Construction of an approximate solution sequence
We will use the Kato method [10] to construct the approximate solutions.We introduce the following approximate system of ( QG ), for k ∈ N ,( QG ) k ( ∂ t θ − k ∆ θ + ( − ∆) α θ + u θ . ∇ θ = 0 θ k (0 , x ) = θ ( x ) . The existence of solution of ( QG ) k based on fixed point theorem.( QG ) k has following integral form θ k = e t ( k ∆ − ( − ∆) α ) θ − Z t e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ ) dτ Let X T = n u ∈ C (cid:16) [0 , T ]; H sa,α − (cid:0) R (cid:1)(cid:17) ; k u k L ∞ T ( H sa,α − ) . ≤ k θ k H sa,α − o and we consider F : X T → C ([0 , T ]; H sa,α − ( R )) θ e t ( k ∆ − ( − ∆) α ) θ − Z t e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ ) dτ • Firstly, we prove F ( X T ) ⊂ X T , we have k e t ( k ∆ − ( − ∆) α ) θ k H sa,α − ≤ k θ k H sa,α − . To estimate the non linear part, we can write: k Z t e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ ) dτ k H sa,α − ≤ Z t k e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ ) k H sa,α − dτ. SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 9
We have k e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ ) k H sa,α − ≤ Z (cid:0) | ξ | (cid:1) s e a | ξ | α (cid:12)(cid:12)(cid:12) F (cid:16) e ( t − τ )( k ∆ − ( − ∆) α ) div( θu θ ) (cid:17) ( τ, ξ ) (cid:12)(cid:12)(cid:12) dξ ≤ Z (cid:0) | ξ | (cid:1) s e a | ξ | α e − t − τ )( k | ξ | + | ξ | α ) | ξ | | d θu θ | dξ ≤ Z (cid:0) | ξ | (cid:1) s e a | ξ | α e − t − τ ) k | ξ | t − τ ) 1 k | ξ | k t − τ ) | d θu θ | dξ ≤ k t − τ ) k θu θ k H sa,α − ≤ C ( a, s, α ) k t − τ ) k u θ k H sa,α − k θ k H sa,α − , ( s > , then k Z t e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ ) dτ k H sa,α − ≤ Z t C s (cid:18) k t − τ ) (cid:19) k θ k H sa,α − dτ ≤ C ( a, s, α ) √ k √ T k θ k L ∞ T ( H sa,α − ) . Then, for
T < k θ k H sa,α − + C s √ k √ T k θ k L ∞ T ( H sa,α − ) ≤ k θ k H sa,α − . We get F ( X T ) ⊂ X T . • We show that F is contracting with an additional condition over T .We have k F ( θ )( t ) − F ( θ )( t ) k H sa,α − ≤ k Z T e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ θ − u θ . ∇ θ ) dτ k H sa,α − ≤ k Z T e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ . ∇ ( θ − θ ) + ( u θ − u θ ) . ∇ θ ) dτ k H sa,α − ≤ K + K , with K = k Z T e ( t − τ )( k ∆ − ( − ∆) α ) u θ . ∇ ( θ − θ ) dτ k H sa,α − ,K = k Z T e ( t − τ )( k ∆ − ( − ∆) α ) ( u θ − u θ ) . ∇ θ dτ k H sa,α − . Since θ , θ ∈ X R,T and k u θ k H sa,α − = k θ k H sa,α − , we obtain ( K ≤ C s √ k √ T k θ − θ k H sa,α − K ≤ C s √ k √ T k θ − θ k H sa,α − Then k F ( θ )( t ) − F ( θ )( t ) k H sa,α − ≤ C s √ k √ T k θ − θ k L ∞ T ( H sa,α − ) We choose
T > C s √ k √ T < F .For these choices, fixed point theorem guarantee the existence unique solution θ k of ( QG ) k in C ([0 , T ] , H sa,α − ) . Step2: Energy estimates and uniform time
Let θ k ∈ C ([0 , T ∗ k ) , H sa,α − ) the maximal solution of ( QG ) k , given by the first step with T ∗ k ∈ (0 , ∞ ]. Taking the scalar product in H sa,α − and taking into account of (2.3), weobtain12 ddt k θ k k H sa,α − + 1 k k| D | θ k k H sa,α − + k| D | α θ k k H sa,α − ≤ |h u θ k . ∇ θ k , θ k i| H sa,α − ≤ C k θ k k H sa,α − k| D | α θ k k H sa,α − . By the convex inequality ab ≤ a + b , we have12 ddt k θ k k H sa,α − + 1 k k| D | θ k k H sa,α − + 12 k| D | α θ k k H sa,α − ≤ C k θ k k H sa,α − , Integrating on [0 , t ], we obtain k θ k ( t ) k H sa,α − + 1 k Z t k| D | θ k k H sa,α − + Z t k| D | α θ k k H sa,α − ≤ k θ k H sa,α − + C Z t k θ k ( t ) k H sa,α − Let t k > t k = sup { t ≥ / sup ≤ z ≤ t k θ k ( z ) k H sa,α − < k θ k H sa,α − } . By continuity of ( t
7→ k θ k ( t ) k H sa,α − ) we get 0 < t k < T ∗ k .For all 0 ≤ t < t k , we have k θ k ( t ) k H sa,α − ≤ k θ k H sa,α − + Ct k θ k H sa,α − , then k θ k ( t ) k H sa,α − + 1 k Z t k| D | θ k k H sa,α − + Z t k| D | α θ k k H sa,α − ≤ k θ k H sa,α − + 16 Ct k θ k H sa,α − Let
T > T k θ k H sa,α − = 2 :then T = 1 C k θ k H sa,α − . For t ∈ [0 , min( T, t k )) one has k θ k ( t ) k H sa,α − ≤ k θ k H sa,α − . Therefore, by continuity of ( t
7→ k θ ( t ) k H sa,α − ), we get t k > T. To choose the above T one deduce that ( θ k ) is bounded in C ([0 , T ] , H sa,α − ) and ( | D | α θ k )in L ([0 , T ] , H sa,α − ). Moreover, we have, for 0 ≤ t ≤ T k θ k ( t ) k H sa,α − + 1 k Z t k| D | θ k k H sa,α − + Z t k| D | α θ k k H sa,α − ≤ k θ k H sa,α − + 16 Ct k θ k H sa,α − = M t . (2.6)which is necessarily in the last step. Step3: Uniqueness and local existence of solution for ( QG ). Let θ k and θ k ′ twosolutions for ( QG ) k , ( QG ) k ′ respectively with k < k ′ . Put ω k,k ′ = θ k − θ k ′ : ω , we get(2.7) ∂ t ω − k ∆ ω + (cid:18) k − k ′ (cid:19) ∆ θ k ′ + ( − ∆) α ω + u ω . ∇ θ k − u θ k ′ ∇ ω = 0 . SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 11
Taking the product scalar in H s − a,α − with ω in (2.7), we get12 ddt k ω k H s − a,α − + k| D | α ω k H s − a,α − ≤ I + I + I + I , with I = k |h ∆ ω, ω i H s − a,α − | ,I = ( k − k ′ ) |h ∆ θ k ′ , ω i H s − a,α − | ,I = |h u ω . ∇ θ k , ω i H s − a,α − | ,I = |h u θ k ′ . ∇ ω, ω i H s − a,α − | Then we get I ≤ k k∇ ω k H s − a,α − . and I ≤ ( 1 k − k ′ ) k∇ θ k ′ k H s − a,α − k∇ ω k H s − a,α − . Since s > H s − a,α − is algebra, then I ≤ C k ω k H s − a,α − k∇ θ k k H s − a,α − Using the lemma 2.4 (2.4), to estimate I I ≤ C k θ k ′ k H s − a,α − k| D | α ω k H s − a,α − k ω k H s − a,α − ≤ C k θ k H sa,α − . By inequality ab ≤ a + b for I , we obtain12 ddt k ω k H s − a,α − + 12 k| D | α ω k H s − a,α − ≤ k k∇ ω k H s − a,α − + ( 1 k − k ′ ) k θ k ′ k H s − a,α − k∇ ω k H s − a,α − + C k ω k H s − a,α − ≤ C ( 1 k + 1 k ′ ) k θ k H s − a,α − + C k ω k H s − a,α − . Then 12 ddt k ω k H s − a,α − ≤ C ( 1 k + 1 k ′ ) k θ k H s − a,α − + C k ω k H s − a,α − Using the Gronwall’s lemma to obtain for 0 < t < T k ω ( t ) k H s − a,α − ≤ Ct ( 1 k + 1 k ′ ) k θ k H s − a,α − exp( Ct ) ≤ CT ( 1 k + 1 k ′ ) k θ k H s − a,α − exp( C ′ ) −→ if k, k ′ → ∞ . Thus, lim k →∞ θ k exists: θ ∈ C (cid:16) [0 , T ]; H s − a,α − (cid:17) .Moreover, we have k ω ( t ) k H s − a,α − + Z T k| D | α ω k H s − a,α − ≤ CT ( 1 k + 1 k ′ ) k θ k H s − a,α − exp( C ′ k θ k H s − a,α − ) −→ when k, k ′ → ∞ . Which implies that ( | D | α θ k ) is a Cauchy sequence in the Hilbert space L ([0 , T ] , H s − a,α − ),then by uniqueness of limit, we get | D | α θ ∈ L ([0 , T ] , H s − a,α − ). The uniqueness of solution
Suppose we have two solutions of ( QG ) θ and θ .Let ω = θ − θ in C (cid:16) [0 , T ]; H s − a,α − (cid:17) then ω (0) = 0 and ∂ t ω + ( − ∆) α ω + δ. ∇ θ − u θ . ∇ ω = 0 . Taking the scalar product in L in the last equation with ω we thus get12 ddt k ω k L + k| D | α ω k L ≤ |h ω. ∇ θ , ω i| L + |h u θ . ∇ ω, ω i| L ≤ C k∇ θ k L k ω k L ≤ C k∇ θ k H s − a,α − k ω k L . By inequality ab ≤ a + b , we get ddt k ω k L + 2 k| D | α ω k L ≤ C k ω k L Since ω (0) = 0 Gronwall Lemma gives us the desired result. Step4: Continuity in time of the solution. • We have θ k ⇀ θ uniformly in H sa,α − . Indeed:Since θ k → θ in H s − a,α − then θ k ⇀ θ in H s − a,α − , from remark (2.5) it follows thatsup ≤ t ≤ T |h θ k ( t ) /f i H s − a,α − − h θ ( t ) /f i H s − a,α − | → , ∀ f ∈ H s − a,α − ( R )sup ≤ t ≤ T |h θ k ( t ) /g i H s − a,α − − h θ ( t ) /g i H s − a,α − | → , ∀ g ∈ e − a | D | α S ( R )Let ϕ = F − ( | ξ | b g ( ξ ))sup ≤ t ≤ T |h θ k ( t ) /ϕ i H sa,α − − h θ ( t ) /ϕ i H sa,α − | → , ∀ ϕ ∈ e − a | D | α S ( R )sup ≤ t ≤ T |h θ k ( t ) /ϕ i H sa,α − − h θ ( t ) /ϕ i H sa,α − | → , ∀ ϕ ∈ H sa,α − ( R ) . • We have θ ∈ L ∞ (cid:16) [0 , T ]; H sa,α − (cid:17) . Indeed: Let ϕ ∈ H sa,α − such that k ϕ k H sa,α − ≤ h θ ( t ) /ϕ i H sa,α − = h θ ( t ) − θ k ( t ) /ϕ i H sa,α − + h θ k ( t ) /ϕ i H sa,α − then |h θ ( t ) /ϕ i H sa,α − | ≤ |h θ ( t ) − θ k ( t ) /ϕ i H sa,α − | + |h θ k ( t ) /ϕ i H sa,α − |≤ sup ≤ z ≤ T |h θ ( z ) − θ k ( z ) /ϕ i H sa,α − | + sup ≤ z ≤ T |h θ k ( t ) /ϕ i H sa,α − |≤ sup ≤ z ≤ T |h θ ( z ) − θ k ( z ) /ϕ i H sa,α − | + 2 k θ k H sa,α − k → ∞ implies |h θ ( t ) /ϕ i H sa,α − | ≤ k θ k H sa,α − then sup k ϕ k Hsa,α − ≤ |h θ ( t ) /ϕ i H sa,α − | = k θ ( t ) k H sa,α − ≤ k θ k H sa,α − . SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 13 • θ is solution of ( QG ). In fact ∂ t θ k ⇀ ∂ t θ. | D | α θ k ⇀ | D | α θ.u θ k . ∇ θ k ⇀ u θ . ∇ θ, in fact k u θ k . ∇ θ k − u θ . ∇ θ k H s − a,α − ≤ k ( u θ k − u θ ) θ k − u θ ( θ k − θ ) k H s − a,α − ≤ k ( u θ k − u θ ) θ k k H s − a,α − + k u θ ( θ k − θ ) k H s − a,α − ≤ k ( u θ k − u θ ) k H s − a,α − k θ k H sa,α − → . • θ is weakly continuous in H sa,α − . Indeed, let ϕ ∈ H sa,α − and 0 < t ≤ t ′ ≤ T , we have |h θ ( t ) − θ ( t ′ ) /ϕ i H sa,α − | ≤ |h θ ( t ) − θ k ( t ) + θ k ( t ) − θ k ( t ′ ) + θ k ( t ′ ) − θ ( t ′ ) /ϕ i H sa,α − |≤ ≤ z ≤ T |h θ ( z ) − θ k ( z ) /ϕ i H sa,α − | + |h θ k ( t ) − θ k ( t ′ ) /ϕ i H sa,α − | → t t ′ . • θ is right-continuous at t . Indeed, Since θ is weakly continuous in H sa,α − it suffices toprove lim sup t → + k θ ( t ) k H sa,α − ≤ k θ k H sa,α − For t > θ k ( t ) → θ ( t )) in H s − a,α − implies the existence a subsequence of ( θ k )such that b θ ϕ ( k ) → b θ a.e For (2.6), we have k θ ϕ t ( k ) ( t ) k H sa,α − ≤ M t . Fatou’s Lemma gives k θ ( t ) k H sa,α − ≤ lim inf k θ α ( k ) ( t ) k H sa,ϕt ( k ) − . Then k θ ( t ) k H sa,α − ≤ M t andlim sup t → + k θ ( t ) k H sa,α − ≤ k θ k H sa,α − . In sum, θ is right-continuous at 0.To prove the same result at any t ∈ [0 , T ], we consider the follow system( QG ) ( ∂ t γ + ( − ∆) α γ + u γ . ∇ γ = 0 γ (0 , x ) = θ ( t , x ) . By the result just proved, γ is continuous at t = 0 by the uniqueness of solution, we obtainthe right-continuous at t of θ . • θ is left-continuous at t . Indeed, For 0 < t ≤ t and 1 < s k < s such that s k ր s . Onehas k θ ( t ) k H ska,α − + 12 Z t t k| D | α θ k H ska,α − = k θ ( t ) k H ska,α − − Z t t h u θ . ∇ θ, θ i H ska,α − | {z } I But | I | ≤ s k + α − ( s k + aα ) Z t t k θ k H ska,α − k| D | α θ k H ska,α − ≤ s + α − ( s + aα ) Z t t k θ k H ska,α − k| D | α θ k H ska,α − Then | I | ≤ C ( t − t ) k θ k H ska,α − + 12 Z t t k| D | α θ k H ska,α − Then k θ ( t ) k H ska,α − ≤ k θ ( t ) k H ska,α − + C ( t − t ) k θ k H ska,α − + Z t t k| D | α θ k H ska,α − ≤ k θ ( t ) k H sa,α − + C ( t − t ) k θ k H sa,α − + Z t t k| D | α θ k H sa,α − . By Monotone convergence theorem, we get k θ ( t ) k H sa,α − ≤ k θ ( t ) k H sa,α − + C ( t − t ) k θ k H sa,α − + Z t t k| D | α θ k H sa,α − Since, | D | α θ is bounded in L ([0 , T ] , H sa,α − ). Then, we pass lim sup t t we obtain k θ ( t ) k H sa,α − ≤ k θ ( t ) k H sa,α − . Then, by proposition 2.1, we get the left continuous at t which complete the proof oftheorem.2.4. Exponential type explosion.Theorem 2.8.
Let θ ∈ H sa,α − , then there exists a unique maximal solution in C ([0 , T ∗ ) , H sa,α − ) ,with T ∗ ∈ (0 , ∞ ] .Moreover, if T ∗ < ∞ then lim sup t ր T ∗ k θ ( t ) k H sa,α − = ∞ . (2.8) Proof.
This proof is inspired from the work of Nader Masmoudi (see [8]).By the local existence step of the proof of theorem 1.2, we get θ ∈ C ([0 , T ] , H sa,α − ) , with T = C k θ k H sa,α − . Consider the following system( P ) ( ∂ t γ + | D | α γ + u γ . ∇ γ = 0 γ (0 , x ) = θ ( T ) ∈ H sa,α − . Then there exist unique solution γ ∈ C ([0 , T ] , H sa,α − ) of ( P ) such that: T = C k γ k H sa,α − = C | θ ( T ) k H sa,α − . SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 15
By the uniqueness of solution we get: γ ( t ) = θ ( T + t ); ∀ ≤ t ≤ T then θ ∈ C ([0 , T + T ] , H sa,α − ) such that T + T < T ∗ . Then we can obtain T , T , T , .., T n such that T k = C k θ ( T + .. + T k − ) k H sa,α − ; ∀ ≤ k ≤ n. Consider the following system( P k ) ( ∂ t δ + | D | α δ + u δ . ∇ δ = 0 δ (0 , x ) = θ ( P nk =1 T k ) ∈ H sa,α − . Then there is a unique solution of ( P k ), δ ∈ C ([0 , T n +1 ] , H sa,α − ).with T n +1 = C k θ ( P nk =1 T k ) k H sa,α − . Also, by uniqueness, we get δ ( t ) = θ ( n X i =1 T i + t ); ∀ ≤ t ≤ T n +1 . Then T + .. + T n +1 < T ∗ , which gives that P ∞ k =1 T k ≤ T ∗ . Then T k −→ k → + ∞ . Thus C k θ ( P ki =1 T i ) k H sa,α − −→ k →∞ , which implies that k θ ( k X i =1 T i ) k H sa,α − −→ k →∞ ∞ . Then P ∞ i =1 T i = T ∗ and lim sup t ր T ∗ k θ ( t ) k H sa,α − = + ∞ . Proposition 2.9.
Let s > , a > and < α < / . If θ is a maximal solution of ( QG ) in C ([0 , T ∗ ) , H sa,α − ( R )) with T ∗ is finite, then Z T ∗ t kF ( e aα | D | α ∇ θ ( τ )) k L dτ = ∞ , ∀ ≤ t < T ∗ . (2.9) C ( a, s ) T ∗ − t ≤ k θ ( t ) k H sa,α − , ∀ t ∈ [0 , T ∗ ) . (2.10) Proof.
Clearly, by theorem 2.8, lim t → T ∗ k θ ( t ) k H sa,α − = ∞ . Proof of 2.9. Taking the scalar product with θ in H sa,α − ( R ) and using (2.2), we obtain12 ddt k θ ( t ) k H sa,α − + k| D | α θ ( t ) k H sa,α − ≤ |h u θ . ∇ θ, θ i H sa,α − |≤ C kF ( e aα | D | α ∇ θ )( t ) k L k| D | α θ k H sa,α − k θ k H sa,α − , with C = C ( a, s, α ) = 2 s + α − ( s + aα ) From inequality xy ≤ x + y and integrating on[ t, T ] ⊂ [0 , T ∗ ) we get k θ ( T ) k H sa,α − + Z Tt k| D | α θ ( t ) k H sa,α − ≤ k θ ( t ) k H sa,α − + Z Tt kF ( e aα | D | α ∇ θ )( τ ) k L k θ k H sa,α − dτ with C ( z, s ) is increasing with respect to z then C ( aα, s ) ≤ C ( a, s ).Gronwall’s Lemma implies k θ ( T ) k H sa,α − + Z Tt k| D | α θ ( τ ) k H sa,α − ≤ C k θ ( t ) k H sa,α − exp( C Z Tt kF ( e aα | D | α ∇ θ )( τ ) k L )The fact that the fact that lim sup T ր T ∗ k θ ( t ) k H sa,α − = ∞ implies Z T ∗ t kF ( e aα | D | α ∇ θ )( t ) k L dτ = ∞ , ∀ ≤ t < T ∗ . and lim sup T ր T ∗ kF ( e aα | D | α ∇ θ )( t ) k L = ∞ . Proof of 2.10. Due to (2.7) and using inequality xy ≤ x + y ,with x = k| D | α θ ( t ) k H sa,α − and y = C k θ ( t ) k H sa,α − , we get12 ddt k θ ( t ) k H sa,α − + k| D | α θ ( t ) k H sa,α − ≤ k| D | α θ ( t ) k H sa,α − k θ ( t ) k H sa,α − Integrating on [ t , t ] ⊂ [0 , T ∗ ) and applying Gronwall’s Lemma we get k θ ( t ) k H sa,α − + Z tt k| D | α θ ( τ ) k H sa,α − ≤ k θ ( t ) k H sa,α − exp( C Z tt k θ ( τ ) k H sa,α − ) . Then k θ ( t ) k H sa,α − exp( − C Z tt k θ ( τ ) k H sa,α − ) ≤ k θ ( t ) k H sa,α − . Integrating over [ t , T ] ⊂ [0 , T ∗ ) we obtain1 − exp( − C Z Tt k θ ( τ ) k H sa,α − ) ≤ C k θ ( t ) k H sa,α − ( T − t ) . Taking T → T ∗ , we get1 − exp( − C Z T ∗ t k θ ( τ ) k H sa,α − ) ≤ C k θ ( t ) k H sa,α − ( T ∗ − t ) . But, from Cauchy-Schwartz inequality, we have kF ( e aα | D | α ∇ θ )( t ) k L ≤ C ( a, s, α ) k θ ( t ) k H sa,α − SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 17 with C = C ( a, s, α ) = Z e − a (1 − α ) | ξ | α | ξ | (1 + | ξ | ) − s dξ < ∞ According this with Z T ∗ t kF ( e aα | D | α ∇ θ )( τ ) k L dτ = ∞ , ∀ ≤ t < T ∗ . we obtain the desired result and Proof of proposition is finished. Proof of (1.1).
This proof is done in two steps: • Step1 we prove C ( a, s ) T ∗ − t ≤ k θ ( t ) k H s Since 0 < a √ α ≤ a we have the following embedding H sa,α − ֒ → H sa √ α,α − which impliesthat θ ∈ C ([0 , T ∗ ) , H sa √ α,α − ) and if T ∗ a √ α,α − the maximal time of existence some solutionin H sa √ α,α − we get T ∗ ≤ T ∗ aα,α − . (2.11)The fact that lim sup T ր T ∗ kF ( e aα | D | α ∇ θ )( t ) k L = ∞ implies lim sup t → T ∗ k θ ( t ) k H saα,α − = ∞ (2.12)Indeed, From Cauchy-Schwartz, we have kF ( e aα | D | α ∇ θ ( τ )) k L = Z ξ e − a ( √ α − α ) | ξ | α | ξ | (1 + | ξ | ) − s (1 + | ξ | ) s e √ α | ξ | α | b θ ( ξ ) | dξ ≤ C k θ ( t ) k H sa √ α,α − Combining (2.11) and (2.12) we obtain T ∗ = T ∗ a √ α,α − . Therefore, C ( a, s ) T ∗ − t ≤ k θ ( t ) k H sa √ α,α − , ∀ t ∈ [0 , T ∗ ) . The same principle gives us T ∗ = T ∗ aα ,α − = ... = T ∗ aα n ,α − . By induction, we can conclude that, for any n ∈ N .C ( a, s ) T ∗ − t ≤ k θ ( t ) k H saα n ,α − Thanks to the convergence dominated theorem, we get C ( a, s ) T ∗ − t ≤ k θ ( t ) k H s As lim k t → T ∗ θ ( t ) k H s = ∞ and k θ ( t ) k L ≤ k θ k L , ∀ t ∈ [0 , T ∗ ) . Then, there is a time T such that k θ ( t ) k H s ≥ k θ k L ; ∀ T ≤ t < T ∗ ) . Then C/ T ∗ − t ≤ k θ ( t ) k H s ; ∀ T ≤ t < T ∗ . (2.13)But, we know that k θ ( t ) k H sa,α − = X k ≥ (2 a ) k k ! k θ ( t ) k H s + kα • Step2 , we prove the exponential type explosion.Using the energy estimate and interpolation for 0 < s < s + kα we obtain k θ ( t ) k ˙ H s ≤ (2 π ) k θ ( t ) k kα s + kα L k θ ( t ) k s s + kα ˙ H s + kα k θ ( t ) k ˙ H s ≤ (2 π ) k θ k kα s + kα L k θ ( t ) k s s + kα ˙ H s + kα . From (2.13), we get C/ T ∗ − t ≤ (2 π ) k θ k kα s + kα L k θ ( t ) k s s + kα ˙ H s + kα , and ( C/ T ∗ − t ) s + kαs ≤ (2 π ) k θ k kαs L k θ ( t ) k H s + kα , which implies C ( T ∗ − t ) . ( C T ∗ − t ) kαs ≤ k θ ( t ) k H s + kα With ( C = ( C ) s C = C π ) k θ k L Multiplying both sides by (2 a ) k k ! we thus get C ( T ∗ − t ) . k ! ( 2 aC ( T ∗ − t ) ) kαs ≤ (2 a ) k k ! Z R | ξ | s | ξ | kα | b θ ( ξ ) | dξ Summing over { k ≥ } to obtain C ( T ∗ − t ) exp( 2 aC ( T ∗ − t ) αs ) ≤ k θ ( t ) k H sa,α − . Which complete the proof. 3.
Global Solution
In this section, we give the proof of theorem 1.3. Let θ ∈ C ((0 , T ∗ ) , H sa,α − ) be a globalsolution of ( QG ). • We start by proving that if T ∗ < ∞ , then Z T ∗ k| D | α θ k H sa,α − dτ = ∞ . SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 19
Assume that there exists
M > R T ∗ k| D | α θ k H sa,α − dτ ≤ M For a select time T ∈ (0 , T ∗ ) such that R T ∗ T k| D | α θ k H − α dτ < . Then, for all t ∈ [ T , T ∗ ) k θ ( t ) k H sa,α − + Z tT k θ k H sa,α − dτ ≤ k θ ( T ) k H sa,α − + C Z tT k| D | α θ ( τ ) k H sa,α − k| D | α θ k H sa,α − dτ ≤ k θ ( T ) k H sa,α − + C sup z ∈ [ T ,t ] k θ ( z ) k H sa,α − Z tT k| D | α θ k H sa,α − dτ ≤ k θ ( T ) k H sa,α − + C sup z ∈ [ T ,t ] k θ ( z ) k H sa,α − ( p t − T Z tT k| D | α θ k H sa,α − dτ ) / ≤ k θ ( T ) k H sa,α − + ( sup z ∈ [ T ,t ] k θ ( z ) k H sa,α − C p T ∗ − T √ M ) . We can suppose that C p T ∗ − T √ M < , Thus, we get sup z ∈ [ T ,T ∗ ] k θ ( t ) k H − α ≤ k θ ( T ) k H sa,α − . Put M = max( max z ∈ [0 ,T ] k θ ( z ) k H sa,α − ; √ k θ ( T ) k H sa,α − ) , we get k θ ( t ) k H sa,α − ≤ M which contradicts with (2.8). • Now, we show that if θ ∈ H sa,α − ( R ) such that k θ k H sa,α − < C , we get a globalsolution in C b ( R + , H sa,α − ( R )).Combining (2.5) and the convex inequality ab ≤ a + b we get k θ ( t ) k H sa,α − + Z t k| D | α θ k H sa,α − dτ ≤ k θ k H sa,α − + C Z t k θ ( τ ) k H sa,α − k| D | α θ ( τ ) k H sa,α − dτ Let T = sup ≤ t ≤ T ∗ { sup ≤ z ≤ t k θ ( z ) k H sa,α − < k θ k H sa,α − } By continuity of ( t → k θ ( t ) k H sa,α − ) we get T > ≤ t < T , we have k θ ( t ) k H sa,α − + Z t k| D | α θ k H sa,α − dτ ≤ k θ k H sa,α − . This implies T = T ∗ , and Z T ∗ k| D | α θ k H sa,α − dτ < ∞ . Hence T ∗ = ∞ and k θ ( t ) k H sa,α − + Z t k| D | α θ k H sa,α − dτ ≤ k θ k H sa,α − , ∀ t ≥ . Long time decay of global solution.
In this section, we prove if θ ∈ C ( R + , H sa,α − ( R ))is a global solution of ( QG ) then k θ ( t ) k H sa,α − decays to zero as time goes to infinity. Asa first step for proving the theorem 1.4, let us show the following proposition: Proposition 3.1. If θ is a solution of ( QG ) such that θ ∈ C b ( R + , H sa,α − ( R )) and | D | α θ ∈ L ( R + , H sa,α − ( R )) , then lim t →∞ k θ ( t ) k H sa,α − = 0 . Proof.
We recall the following L energy estimate(3.1) k θ k L + 2 Z t k| D | α θ k L ≤ k θ k L . As 0 ≤ α ≤ s , then | ξ | s ≤ (cid:26) | ξ | α if | ξ | < | ξ | s + α if | ξ | > ≤ (1 + | ξ | ) s | ξ | α . By the above inequality we obtain k θ ( t ) k H sa,α − = Z R | ξ | s e a | ξ | α | b θ ( ξ ) | dξ ≤ Z {| ξ | < } | ξ | s e a | ξ | α | b θ ( ξ ) | dξ + Z {| ξ | > } | ξ | s e a | ξ | α | b θ ( ξ ) | dξ ≤ e a k θ ( t ) k H α + k| D | α θ ( t ) k H sa,α − Combining (3.1) and the fact | D | α θ ∈ L ( R + , H sa,α − ( R )), we obtain θ ∈ L ( R + , ˙ H sa,α − ) . Let ε > A ε = { t ≥ k θ ( t ) k H sa,α − > ε } . We have M : e a k θ k L + k θ k L ≥ Z R k θ ( t ) k H sa,α − ≥ Z A ε k θ ( t ) k H sa,α − ≥ λ ( A ε ) ε Let T ε = ε − M then λ ( A ε ) ≤ T ε and there is a t ∈ [0 , T ε + 2] \ A ε which implies k θ ( t ) k H sa,α − < ε. (3.2)Let us consider the following equation( QG ) ( ∂ t γ + | D | α γ + u γ . ∇ γ = 0 γ (0) = θ ( t ) . The existence and uniqueness of a solution to the quasi geostrophic equation gives for all t > , γ ( t ) = θ ( t + t ) then k θ ( t + t ) k H sa,α − + Z t k| D | α θ ( t + τ ) k H sa,α − dτ ≤ k θ ( t ) k H sa,α − < ε , ∀ t > , which completes the proof of proposition. Proof theorem 1.4.
By the embedding H sa,α − ֒ → H saα,α − we get θ ∈ C ( R + , H saα,α − ( R )) SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 21
Similarly, we get for k ∈ N θ ∈ C ( R + , H saα k ,α − ( R ))On the other hand, (see the appendix), it is shown thatlim t →∞ k θ ( t ) k H s = 0 , which implies for any small enough ε > t > k θ ( t ) k H s < ε, t ≥ t . But the Dominated Convergence Theorem yields lim k →∞ k θ ( t ) k H saαk,α − = k θ ( t ) k H s .Hence, there exists k > k ≥ k we have k θ ( t ) k H saαk,α − < ε. Using proposition 3.1, we get k θ ( t + t ) k H saαk ,α − ≤ ε, ∀ t ≥ , and k θ ( t + t ) k H saαk ,α − + Z t k| D | α θ ( t + τ ) k H saαk ,α − dτ ≤ k θ ( t ) k H saαk ,α − < ε , ∀ t > , It follows from (2.2) of Lemma 2.4 that |h u θ . ∇ θ, θ i H sa,α − | ≤ C k e aα | D | α θ k X k| D | α θ k H sa,α − k θ k H sa,α − ≤ CC k| D | α θ k H saα,α − k| D | α θ k H sa,α − k θ k H sa,α − with C = Z | ξ | (1 + | ξ | ) s | ξ | α = π Z ∞ r (1 + r ) s r α dr< ∞ (2 s − α > CC is independent of a .In order to bound θ ( t ) in H sa,α − ( R ), we take the scalar product with θ in H saα k − ,α − toget12 ddt k θ ( t ) k H saαk − ,α − + k| D | α θ k H saαk − ,α − ≤ C k| D | α θ k H saαk ,α − k| D | α θ k H saαk − ,α − k θ k H saαk − ,α − ≤ k| D | α θ k H saαk − ,α − + C k θ k H saαk − ,α − k| D | α θ k H saαk ,α − Integrating over [ t , t ] we obtain k θ ( t ) k H saαk − ,α − + Z tt k| D | α θ k H saαk − ,α − dτ ≤ k θ ( t ) k H saαk − ,α − + C Z tt k| D | α θ ( τ ) k H saαk ,α − k θ ( τ ) k H saαk − ,α − dτ By Gronwall’s Lemma we get k θ ( t ) k H saαk − ,α − + Z tt k| D | α θ k H saαk − ,α − dτ ≤ k θ ( t ) k H saαk − ,α − exp (cid:0) C Z tt k| D | α θ ( τ ) k H saαk ,α − (cid:1) ≤ k θ ( t ) k H saαk − ,α − exp (cid:0) C Z ∞ t k| D | α θ ( τ ) k H saαk ,α − (cid:1) ≤ k θ ( t ) k H saαk − ,α − exp (cid:0) C Z ∞ k| D | α θ ( τ ) k H saαk ,α − (cid:1) ≤ C k . Then k θ ( t ) k H saαk − ,α − = M k , ∀ t ≥ , with M k = max( k θ k L ∞ ([0 ,t ] ,H saαk − ,α − ) ,C k ) . Similarly, by finite and decreasing induction, we can deduce that θ ( t ) is bounded in H sa,α − ( R ) , t ≥ Appendix
In this section, we prove that lim t →∞ k θ ( t ) k H s = 0 , if θ ∈ C ( R + , H sa,α − ( R )) is a global solution of ( QG ).In [2], we have ( lim t →∞ k θ ( t ) k ˙ H − α = 0 , R ∞ k| D | α θ k H − α < ∞ . Using Lemma 2.3, we get for all t > t k θ ( t ) k H s + 2 Z tt k| D | α θ k H s dτ ≤ k θ ( t ) k H s + 2 C Z tt k θ ( τ ) k ˙ H − α k| D | α θ ( τ ) k H s dτ Let t > k θ ( t ) k ˙ H − α < C , t ≥ t . Then k θ ( t ) k H s + Z tt k| D | α θ k H s dτ ≤ k θ ( t ) k H s Let t > k θ ( t ) k ˙ H − α < C , t ≥ t . Which implies C ( R + , H s ( R )) ∩ L ( R + , ˙ H s + α ( R )) . Let k ∈ N , k ≥
2, such that2 − α + k α ≤ s < − α + ( k + 1) α. First case: If s = 2 − α + k α : we want to prove that ( lim t →∞ k θ ( t ) k ˙ H − α + iα = 0 , R ∞ k| D | α θ k H − α + iα < ∞ , ∀ ≤ i ≤ k . SYMPTOTIC STUDY OF A GLOBAL SOLUTION OF SUPER-CRITICAL ... 23
Suppose that lim t →∞ k θ ( t ) k ˙ H − α + iα = 0 , ≤ ∀ i ≤ k − t →∞ k θ ( t ) k ˙ H − α +( i +1) α =0. Let t > k θ ( t ) k ˙ H − α < C α , t ≥ t . As R ∞ k| D | α θ k H − α + iα < ∞ we have Z ∞ t k| D | α θ k H − α + iα < ∞ , which implies that there is t > t such that k θ ( t ) k H − α +( i +1) α < ε. For t ≥ t we get k θ ( t ) k H − α +( i +1) α +2 Z tt k| D | α θ k H − α +( i +1) α ≤ k θ ( t ) k H s +2 C α Z tt k θ k ˙ H − α k| D | α θ k H − α +( i +1) α . Then k θ ( t ) k H − α +( i +1) α + Z tt k| D | α θ k H − α +( i +1) α ≤ k θ ( t ) k H − α +( i +1) α < ε , ∀ t ≥ t . Thus, we get for all k > ( lim t →∞ k θ ( t ) k ˙ H − α + k α = 0 , R ∞ k| D | α θ k H − α + k α < ∞ . Second case:
If 2 − α + k α < s < − α + ( k + 1) α Let ε > r ≥ k θ ( t ) k ˙ H − α < min( 14 C s , ε/ , t ≥ r . For t > a > r , we have k θ ( t ) k H s + 2 Z ta k| D | α θ k H s dτ ≤ k θ ( t ) k H s + 2 C Z ta k θ ( τ ) k ˙ H − α k| D | α θ ( τ ) k H s dτ Then k θ ( t ) k H s + Z ta k| D | α θ k H s dτ ≤ k θ ( a ) k H s . It suffice to prove that, there is a ≥ t such that k θ ( a ) k H s ≤ ε. By interpolation, we obtain k θ ( t ) k H s ≤ k θ ( t ) k − rH − α + k α k θ ( t ) k rH − α +( k α , with r ∈ [0 , k θ ( t ) k − r H s ≤ k θ ( t ) k r − r H − α + k α k θ ( t ) k H − α +( k α . Let ε > E ε = { t ≥ r , k θ ( t ) k ˙ H s > ε } we get M ≥ Z E ε k θ ( t ) k − r ˙ H s dt ≥ ε − r λ ( E ε ) . Then λ ( E ε ) ≤ t ε = ε − r M . For η >
0, there exists a ∈ [ t , t ε + η ] such that a / ∈ E ε andit results that k θ ( a ) k ˙ H s ≤ ε, which prove the desired result. Acknowledgements.
It is pleasure to thank J. Benameur for insightful comments andassistance through this work.
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Department of Mathematics, Faculty of Science of Gab`es, Research Laboratory Mathe-matics and Applications LR17ES11; Tunisia
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