aa r X i v : . [ m a t h . A P ] A ug Uniqueness of the Leray-Hopf solution for a dyadic model
N. D. Filonov
Abstract
The dyadic problem ˙ u n + λ n u n − λ βn u n − + λ β ( n +1) u n u n +1 = 0 with ”smooth” initialdata is considered. The uniqueness of the Leray-Hopf solution is proved. Introduction
We consider the following problem ( ˙ u n ( t ) + λ n u n ( t ) − λ βn u n − ( t ) + λ β ( n +1) u n ( t ) u n +1 ( t ) = 0 , t ∈ [0 , ∞ ) ,u n (0) = a n , n = 1 , , . . . . (0.1)Here u ≡ λ > β > a = { a n } ∈ l . Last decade, many authors pay attention to theproblems of such kind, see [1, 2, 3, 4, 6, 7, 8, 9, 11, 12]. An important feature of the system(0.1) is that it is similar to the system of the Navier-Stokes equations ( ∂ t u − ∆ u + P (( u, ∇ ) u ) = 0 in [0 , ∞ ) × T d , div u = 0 , u | t =0 = a ( x ) . (0.2)Here T d is a d -dimensional torus, and P is the orthoprojector in L ( T d ) onto the subspace ofdivergence-free functions. Both systems can be written in the following abstract way: ( ˙ u + Au + B ( u, u ) = 0 , t ∈ [0 , ∞ ) ,u (0) = a. Here the function u ( t ) takes values in a Hilbert space H , H = l in the case (0.1), and H = { u ∈ L ( T d ) : div u = 0 } in the case (0.2). A is a self-adjoint non-negative unbounded operatorin H . B is a bilinear unbounded map B : H × H → H , having two principal properties: • the cancellation property ( B ( u, u ) , u ) H = 0for the dense set in H of ”good” u ; • an estimate k B ( u, u ) k H C k A σ u k H k A σ u k H holds with some σ , σ . 1he orders σ , σ can vary, but the sum ( σ + σ ) is fixed. For the case of the problem (0.1) wehave σ + σ = β , and for the case (0.2) the imbedding theorems imply σ + σ = d +24 , where d is a dimension of the space variables. Thus, the most interesting value of the parameter β in(0.1) is β = 5 / ( ˙ u n ( t ) − κ n u n − ( t ) + κ n +1 u n ( t ) u n +1 ( t ) = 0 , t ∈ [0 , ∞ ) ,u n (0) = a n , n = 1 , , . . . , (0.3)is also actively studied. It can be obtained from (0.1) by elimination of the linear term λ n u n ( t ); κ = λ β >
1. The system (0.3) is a model for the Euler equations in hydrodynamics (which isthe problem (0.2) without the viscous term ∆ u ). Definition 0.1.
A sequence of functions { u n } is called a weak solution to the problem (0.1) if u n ∈ C ∞ [0 , ∞ ) for all n , and (0.1) is fulfilled. Definition 0.2.
A weak solution { u n } is called a Leray-Hopf solution to the problem (0.1) ifthe inequality ∞ X n =1 (cid:18) u n ( t ) + 2 Z t λ n u n ( τ ) dτ (cid:19) ∞ X n =1 a n ∀ t ∈ [0 , ∞ ) (0.4) holds. In the same manner one defines a Leray-Hopf solution on a finite interval [0 , T ] . It is easy to show that a Leray-Hopf solution exists for any initial data { a n } ∈ l , seeTheorem 2.2 below.We will not introduce a formal definition of a strong solution to (0.1). But we will use thewords ”strong solution” in a non-formal manner, meaning that the functions u n ( t ) decrease fastas n → ∞ in a uniform or in an integral norm. And we would like to warn the reader that thesense may be different in different situations. For example, it is easy to see that if the solution { u n } is strong in such a way that Z T | u n ( t ) | dt = o ( λ − βn ) , n → ∞ , then it is also a Leray-Hopf solution on [0 , T ]. A. Cheskidov proved that the system (0.1) has a global in time strong solution if β
2, andthat a blow-up takes place if β >
3. Let us formulate these results more precisely.
Theorem 0.3 ([4]) . Let β , P ∞ n =1 λ n a n < ∞ . Then the problem (0.1) has a solution { u n } satisfying the relation sup t ∞ X n =1 λ n u n ( t ) < ∞ . heorem 0.4 ([4]) . Let β > , ε > . Then any solution { u n } to (0.1) will have Z T ∞ X n =1 λ ε +1 / βn u n ( t ) ! / dt = + ∞ for some T whenever all a n > and P ∞ n =1 λ βεn a n > M . Here M is a constant depending on ε . For example, one can take a n = 0 for n > , if a is big enough. D. Barbato, F. Morandin and M. Romito have considered the important particular case forthe problem (0.1) when the initial data are non-negative.
Theorem 0.5 ([3]) . Let λ = 2 , < β . Let { a n } ∈ l , a n > . Then there exists a uniqueweak solution { u n } to the problem (0.1) , and u n ( t ) = O ( λ − γn ) , n → ∞ , ∀ γ, ∀ t > . Of course, one would like to remove 1) the condition of non-negativity for the initial data;and 2) the fixation of the value λ = 2. We are unable to prove the existence of a strongsolution in such setting. Instead of it we prove the uniqueness of a Leray-Hopf solution. Weuse heavily the following result of Barbato and Morandin on the uniqueness of a weak solutionto the problem (0.3) if the initial data are non-negative. Theorem 0.6 ([2]) . Let κ > , { a n } ∈ l , a n > . Then there exists a unique weak solution tothe problem (0.3) . This result can be extended to the case of the system (0.1), see Theorem 4.2 below. Usingthis fact, we establish the uniqueness of the Leray-Hopf solution to (0.1) for sufficiently goodinitial data. We formulate now the main result of the paper.
Theorem 0.7.
Let λ > , β > . Assume that { a n } ∈ l if β , and a n = o ( λ (2 − β ) n ) , n → ∞ , if β > . Then the problem (0.1) has a unique Leray-Hopf solution. Remark 0.8.
It will be seen from the proof that we establish a slightely more strong result.Namely, for β > n →∞ | a n | λ ( β − n < ε := min (cid:18) λ + λ β − , λ + λ β − (cid:19) . (0.5)It is easy to show that if there exists a strong solution to (0.1), then it is unique in theclass of the Leray-Hopf solutions, see Theorem 1.4 below. For β
3, we do not know if theLeray-Hopf solution guaranteed by Theorem 0.7 is a strong one. If β > notstrong solution, see Theorem 0.4. The value of the parameter β represents in a sense a ”rate of domination” of the convective (non-linear)term in the equations over the dissipative (linear) term. One can associate such a parameter in an equivalentway with the convective term or with the dissipative one. In [3] the parameter is associated with the convectiveterm, like in (0.1). In [4] it is associated with the linear term:˙ u n + λ αn u n − λ n u n − + λ n +1 u n u n +1 = 0 . The currency exchange is as follows: λ Ch = λ βBMR , λ BMR = λ αCh , αβ = 1 . .4 Stationary solutions We call a stationary solution a sequence { a n } of real numbers satisfying the stationary equations λ n a n − λ βn a n − + λ β ( n +1) a n a n +1 = 0 , a = 0 , n = 1 , , . . . . (0.6) Theorem 0.9 ([3]) . a) Let a sequence { a n } satisfy (0.6) . Then if a n = 0 then a n = 0 for all n n . If a n = 0 , then a n − λ (2 − β ) n − for all n > n . b) Let the parameters β ∈ (2 , and λ be such that λ β − < . Then there exists a non-trivialsolution { a n } to (0.6) such that a n = O (cid:0) λ (2 − β ) n (cid:1) , n → ∞ . The claim a) of this Theorem means that if β l .We establish the existence of stationary solution for all values of the parameters λ and β . Theorem 0.10.
There exists a non-trivial solution { a n } to (0.6) such that • a n = O (cid:0) λ (2 − β ) n (cid:1) , n → ∞ , if β < , • a n = O ( nλ − n ) , n → ∞ , if β = 3 , • a n = O (cid:0) λ − βn/ (cid:1) , n → ∞ , if β > .These estimates can not be improved, as any non-trivial solution { a n } to (0.6) satisfies thefollowing relation: • lim sup n →∞ | a n | λ ( β − n > if β < , • lim sup n →∞ | a n | λ n = + ∞ if β = 3 , • lim sup n →∞ | a n | λ βn/ > if β > . It is clear that a stationary solution is not a Leray-Hopf solution. On the other hand, aLeray-Hopf solution exists for any initial data from l . So, these results mean that a weaksolution to (0.1) may be non-unique if β >
2. Note, that for β ∈ (2 ,
3) we have this non-uniqueness of weak solution with initial data of order a n ∼ λ (2 − β ) n (cf. Theorem 0.7).Note also, that stationary solutions to (0.1) are related to the self-similar solutions to thesystem (0.3) with a blow-up, see [1]. Theorem 0.10 in the case β > In the first section we prove the basic properties of weak solutions and the uniqueness of thestrong solution. In the second section we introduce the notion of Galerkin solution, and describeits properties for ”good” initial data. In the third section we estimate the integral R ∞ u n dt fora non-negative case. In the fourth section we use this estimate to justify the uniqueness ofweak solution in the non-negative case, and then we prove Theorem 0.7. In the last section westudy the steady-state solutions and prove Theorem 0.10.4 .6 Acknowledgements The author thanks T. Shilkin for fruitful discussions.
The content of this subsection is borrowed from the paper [4]. We give the proofs for the sakeof completeness.
Lemma 1.1 ([4]) . Let { u n } be a weak solution to (0.1) .a) If a n > then u n ( t ) > for all t .b) If a n < then either u n ( t ) < for all t , or there exists a time τ n such that u n ( t ) < for t < τ n , u n ( τ n ) = 0 , u n ( t ) > for t > τ n . Proof.
One can rewrite the differential equation for u n as an integral one, u n ( t ) = a n exp (cid:18) − Z t ( λ n + λ β ( n +1) u n +1 ( s )) ds (cid:19) (1.1)+ Z t exp (cid:18) − Z ts ( λ n + λ β ( n +1) u n +1 ( σ )) dσ (cid:19) λ βn u n − ( s ) ds. Therefore, u n ( t ) > a n exp (cid:18) − Z t ( λ n + λ β ( n +1) u n +1 ( s )) ds (cid:19) , and a) follows. Clearly, a) implies b).We introduce the notation l +2 = {{ a n } ∈ l : there is a number M such that a n > n > M } . (1.2) Lemma 1.2 ([4]) . Let { u n } be a weak solution to (0.1) with { a n } ∈ l +2 . Then { u n } is aLeray-Hopf solution.Proof. Let a n > n > M . By virtue of Lemma 1.1 u n ( t ) > n > M . Now, (0.1) impliesfor N > M N X n =1 (cid:0) u n ˙ u n + λ n u n (cid:1) = − λ β ( N +1) u N u N +1 , and therefore, N X n =1 (cid:18) u n ( t ) Z t λ n u n ( τ ) dτ (cid:19) N X n =1 a n ∞ X n =1 a n . Passing to the limit N → ∞ , we get (0.4).It is clear from the proof that the following property of initial data is sufficient: there is asequence n k → ∞ such that a n k >
0. Note also, that a non-trivial stationary solution (whichis not Leray-Hopf) has all a n < n due to Theorem 0.9.5 .2 Uniqueness of the strong solution Lemma 1.3.
Assume that the functions { u n } , { v n } and { w n } satisfy the inequalities | u n ( t ) | Cλ (2 − β ) n , t ∈ [0 , T ] , ∞ X n =1 λ n Z T (cid:0) v n ( t ) + w n ( t ) (cid:1) dt < ∞ . Then the series ∞ X n =1 λ βn Z T | u n ( t ) | ( | v n − ( t ) | + | v n ( t ) | + | v n +1 ( t ) | ) ( | w n − ( t ) | + | w n ( t ) | + | w n +1 ( t ) | ) dt converges. This Lemma is evident.
Theorem 1.4.
Let { u n } and { v n } be the Leray-Hopf solutions to (0.1) corresponding to thesame initial data { a n } ∈ l . Assume that there is a number L such that | u n ( t ) | ε λ (2 − β ) n , for n > L, t ∈ [0 , T ] , where ε = (cid:0) λ + λ β − (cid:1) − . Then u n ≡ v n on [0 , T ] .Proof. We have ddt ( u n v n ) + 2 λ n u n v n = λ βn (cid:0) u n v n − + u n − v n (cid:1) − λ β ( n +1) ( u n v n v n +1 + u n u n +1 v n ) . Therefore, u n ( t ) v n ( t ) − a n + 2 Z t λ n u n v n dτ = Z t (cid:0) λ βn (cid:0) u n v n − + u n − v n (cid:1) − λ β ( n +1) ( u n v n v n +1 + u n u n +1 v n ) (cid:1) dτ, and ∞ X n =1 (cid:18) u n ( t ) v n ( t ) − a n + 2 Z t λ n u n v n dτ (cid:19) = ∞ X n =1 Z t λ β ( n +1) (cid:0) u n +1 v n + u n v n +1 − u n v n v n +1 − u n u n +1 v n (cid:1) dτ (1.3)= ∞ X n =1 Z t λ β ( n +1) ( u n − v n ) ( u n +1 ( u n − v n ) − u n ( u n +1 − v n +1 )) dτ. All series in the last equality converge due to Lemma 1.3.Let us estimate the remainder of the last series. We have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n = L Z t λ β ( n +1) u n +1 ( u n − v n ) dτ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n = L Z t ε λ n +2 ( u n − v n ) dτ, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n = L Z t λ β ( n +1) u n ( u n − v n )( u n +1 − v n +1 ) dτ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n = L Z t ε λ n + β (cid:18) λ ( u n − v n ) + λ u n +1 − v n +1 ) (cid:19) dτ ∞ X n = L Z t ε λ n − β ( u n − v n ) dτ. Now, taking into account the definition of ε and the equality (1.3), we get − ∞ X n =1 (cid:18) u n ( t ) v n ( t ) + 2 Z t λ n u n v n dτ (cid:19) (1.4) − ∞ X n =1 a n + C L X n =1 Z t ( u n − v n ) dτ + 2 ∞ X n = L Z t λ n ( u n − v n ) dτ, where the constant C = 4 λ βL sup n =1 ,...,L k u n k L ∞ λ βL k a k l does not depend on t .Recall that ∞ X n =1 (cid:18) u n ( t ) + 2 Z t λ n u n ( τ ) dτ (cid:19) ∞ X n =1 a n , (1.5) ∞ X n =1 (cid:18) v n ( t ) + 2 Z t λ n v n ( τ ) dτ (cid:19) ∞ X n =1 a n , (1.6)by definition of the Leray-Hopf solutions. Summarizing (1.4) – (1.6) we obtain ∞ X n =1 (cid:18) ( u n ( t ) − v n ( t )) + 2 Z t λ n ( u n ( τ ) − v n ( τ )) dτ (cid:19) C L X n =1 Z t ( u n ( τ ) − v n ( τ )) dτ + 2 ∞ X n = L Z t λ n ( u n ( τ ) − v n ( τ )) dτ, where from ∞ X n =1 ( u n ( t ) − v n ( t )) C Z t L X n =1 ( u n ( τ ) − v n ( τ )) dτ. If we put y ( t ) = P ∞ n =1 ( u n ( t ) − v n ( t )) then the last inequality implies y ( t ) C R t y ( τ ) dτ , andthus, y ( t ) ≡ u n ≡ v n .Note, that this proof is nothing but a simplified version of the standard proof of the unique-ness of a strong solution among Leray-Hopf solutions in the NSE theory. Corollary 1.5.
Let β < . Then the Leray-Hopf solution is unique for arbitrary initial data { a n } ∈ l .Proof. By definition of Leray-Hopf solution we have | u n ( t ) | k a k l for all n and t . Therefore, | u n ( t ) | ε λ (2 − β ) n for sufficiently large n . 7 Galerkin solutions
Galerkin solutions is an important tool for the problem under consideration. In order to defineit let us consider for each N ∈ N the following problem ˙ v ( N ) n + λ n v ( N ) n − λ βn (cid:16) v ( N ) n − (cid:17) + λ β ( n +1) v ( N ) n v ( N ) n +1 = 0 , t ∈ [0 , ∞ ) ,v ( N ) n (0) = a n , n = 1 , . . . , N ; v ( N )0 ≡ , v ( N ) N +1 ≡ . (2.1)It is well known from the ordinary differential equations theory that the problem (2.1) has aunique solution for the small interval of time. The length of this interval depends on the l -normof the initial data (cid:16)P Nn =1 a n (cid:17) / only. Multiplying the equations (2.1) by v ( N ) n and taking thesum over n we get N X n =1 (cid:16) v ( N ) n ˙ v ( N ) n + λ n (cid:0) v ( N ) n (cid:1) (cid:17) = 0 , so 12 N X n =1 (cid:0) v ( N ) n ( t ) − a n (cid:1) + N X n =1 Z t λ n v ( N ) n ( τ ) dτ = 0 , (2.2)and N X n =1 v ( N ) n ( t ) N X n =1 a n . (2.3)Last estimate implies that the problem (2.1) has a global in time solution. Moreover, k v ( N ) n k C [0 ,T ] k a k l , and the equations (2.1) yield the boundedness of the sequence { v ( N ) n } ∞ N = n in C [0 , T ] for each n and for each T . Therefore, there exists a subsequence { v ( N k ) n } converging in C [0 , T ] as N k → ∞ . Applying the diagonal process, we obtain a sequence of numbers { M k } such that v ( M k ) n −→ k →∞ u n in C [0 , T ] for all n, and for all T. Clearly, the limit sequence { u n } satisfies (0.1) on [0 , ∞ ). Definition 2.1.
We call the Galerkin solution to the problem (0.1) a solution { u n ( t ) } con-structed above. Note, that the construction of a Galerkin solution does not imply in general its uniqueness.Now, let us fix M ∈ N . The relation (2.2) yields for N > M M X n =1 (cid:18) v ( N ) n ( t ) + 2 Z t λ n v ( N ) n ( τ ) dτ (cid:19) ∞ X n =1 a n , which implies the same inequality for the limit functions M X n =1 (cid:18) u n ( t ) + 2 Z t λ n u n ( τ ) dτ (cid:19) ∞ X n =1 a n , and therefore, (0.4) holds. Thus, we see that any Galerkin solution to (0.1) is a Leray-Hopfsolution as well. We proved 8 heorem 2.2. Let { a n } ∈ l . Then there is a Leray-Hopf solution to the problem (0.1) . Note, that the constructions of this subsection are also standard.
Theorem 2.3.
Let ε λ − , and a n > − ε λ (2 − β ) n for n > K for some K . Let { u n } be aGalerkin solution to (0.1) . Then u n ( t ) > − ε λ (2 − β ) n for n > K, t ∈ [0 , ∞ ) . Proof.
Let us consider the solution { v ( N ) n } to (2.1) with some N > K . Similarly to (1.1) wehave v ( N ) n ( t ) > a n exp (cid:18) − Z t ( λ n + λ β ( n +1) v ( N ) n +1 ( s )) ds (cid:19) . (2.4)We show now that v ( N ) n ( t ) > − ε λ (2 − β ) n for n = K, . . . , N + 1 , t ∈ [0 , ∞ ) , (2.5)by backward induction in n . For n = N + 1 the claim is clear as v ( N ) N +1 ≡
0. Assume that (2.5)is true for n = k + 1. We have λ k + λ β ( k +1) v ( N ) k +1 ( s ) > λ k (1 − ε λ ) > ε λ − . Now, (2.4) implies that v ( N ) k ( t ) > t if a k >
0, and v ( N ) k ( t ) > a k > − ε λ (2 − β ) k , if a k <
0. Thus, (2.5) is true for n = k .So, (2.5) is proven. Passing in (2.5) to the limit N → ∞ , we obtain the result. Corollary 2.4.
Under the assumptions of Theorem 2.3 we have λ n + λ β ( n +1) u n +1 ( t ) > λ n (1 − ε λ ) , for n > K, t ∈ [0 , ∞ ) . Corollary 2.5.
Under the assumptions of Theorem 2.3, if u n ( t ) for some n > K , then ˙ u n ( t ) > .Proof. From the equation (0.1) we have˙ u n ( t ) = λ βn u n − ( t ) − u n ( t ) (cid:0) λ n + λ β ( n +1) u n +1 ( t ) (cid:1) > u n ( t )
0. 9 .3 Good initial data. Estimate of absolute value
Theorem 2.6.
Let ε (cid:0) λ + λ β − (cid:1) − , and | a n | ε λ (2 − β ) n for n > K for some K . Let { u n } be a Galerkin solution to (0.1) . Assume that there is a number L > K such that | u L ( t ) | ε λ (2 − β ) L for all t ∈ [0 , T ] . Then | u n ( t ) | ε λ (2 − β ) n for all n > L, t ∈ [0 , T ] . (2.6) Proof.
We proceed by induction in n . The case n = L is given by the assumption. Assumethat (2.6) is fulfilled for n = k −
1. By virtue of Corollary 2.4 λ k + λ β ( k +1) u k +1 ( t ) > λ k (1 − ε λ ) > λ k λ − β + 1 . Therefore, by (1.1) and (2.6) for n = k −
1, we have | u k ( t ) | | a k | e − λ ktλ − β +1 + Z t e − λ k ( t − s ) λ − β +1 λ βk ε λ (4 − β )( k − ds = | a k | e − λ ktλ − β +1 + ε λ (2 − β ) k (cid:0) λ + λ β − (cid:1) (cid:18) − e − λ ktλ − β +1 (cid:19) ε λ (2 − β ) k (cid:20) ε (cid:0) λ + λ β − (cid:1) + (cid:0) − ε (cid:0) λ + λ β − (cid:1)(cid:1) e − λ ktλ − β +1 (cid:21) . By assumption, we have 1 − ε (cid:0) λ + λ β − (cid:1) >
0, so, the expression in the rectangle bracketsattains its maximum at t = 0, and this maximum is equal to 1. Therefore, | u k ( t ) | ε λ (2 − β ) k ,and (2.6) is fulfilled for n = k .Theorem 2.6 implies two corollaries. The first one is the strong solvability of the problem(0.1) on a small interval of time. Corollary 2.7.
Let | a n | ε λ (2 − β ) n for n > K , ε (cid:0) λ + λ β − (cid:1) − . Assume moreover, that | a K | < ε λ (2 − β ) K . Let { u n } be a Galerkin solution to (0.1) . Then there is τ > such that | u n ( t ) | ε λ (2 − β ) n for n > K, t ∈ [0 , τ ] . (2.7) Proof.
The function u K is continuous, so there is τ > | u K ( t ) | ε λ (2 − β ) K for t ∈ [0 , τ ] . Now we apply Theorem 2.6 with L = K on the interval [0 , τ ], and we get (2.7) for all n > K .The second corollary of Theorem 2.6 guarantees that the solution is strong whenever thefinal data are non-positive. Corollary 2.8.
Let | a n | ε λ (2 − β ) n for n > K , ε (cid:0) λ + λ β − (cid:1) − . Let { u n } be a Galerkinsolution to (0.1) on [0 , T ] . Assume that { u n ( T ) } / ∈ l +2 , where the set l +2 is defined in (1.2) .Then there is a number L such that | u n ( t ) | ε λ (2 − β ) n for n > L, t ∈ [0 , T ] . Proof.
Choose a number L > K such that u L ( T )
0. By virtue of Lemma 1.1 and Corollary2.5 we have a L u L ( t ) ∈ [ a L ,
0] for all t ∈ [0 , T ] . Therefore, | u L ( t ) | ε λ (2 − β ) L , t ∈ [0 , T ], and the claim follows from Theorem 2.6.10 Estimate of the integral R ∞ u n dt The aim of this section is the following estimate of weak solutions Z ∞ u n ( t ) dt = O ( nλ − βn ) , n → ∞ , for a non-negative case. Such estimate for a finite interval of time is proven in [2] for the system(0.3) under the assumption that the initial data are non-negative, a n > n . We need thisestimate for the system (0.1) and for the initial data from l +2 (all components are non-negativebeginning from some number M ). The proof is very similar to the proof in [2], we give it forthe sake of completeness.We need the following covering lemma, see for example [10, Ch. I]. Lemma 3.1.
Let A be a measurable subset of R . Let I be a family of intervals in R , lengthsof which are bounded. Assume that for all a ∈ A there is an interval ( a, b ) from I . Then wecan select from this family I a disjoint subsequence { ( a k , b k ) } , finite or countable, such that | A | X k ( b k − a k ) . Let { u n } be a weak solution to (0.1) with the initial data { a n } , a n > n > M . By virtueof Lemma 1.2 { u n } is a Leray-Hopf solution, and k{ u n ( t ) }k l k{ a n }k l for all t .Following [2] we introduce the sequence of functions E n ( t ) := n X k =1 u k ( t ) . Then ˙ E n ( t ) + 2 n X k =1 λ k u k ( t ) = − λ β ( n +1) u n ( t ) u n +1 ( t ) . (3.1)In particular, ˙ E n ( t ) n > M − . (3.2)Let us fix y > n > M . We consider a closed set A n ( y ) := { s > u n ( s ) > y > u n +2 ( s ) } . (3.3)For this set we construct a family of intervals I , and then we will apply Lemma 3.1. Let s ∈ A n ( y ). We put t ( s ) := min inf { t > s : u n ( t ) < y/ } , inf { t > s : u n +2 ( t ) > y } ,s + λ n +2 +2 yλ β ( n +2) . (3.4) Lemma 3.2.
Let s ∈ A n ( y ) , and t > s be such that u n ( t ) = y/ . Then E n ( s ) − E n ( t ) > y . roof. By definition of E n ’s E n ( s ) − E n ( t ) = E n − ( s ) − E n − ( t ) + u n ( s ) − u n ( t ) > u n ( s ) − u n ( t ) > y , where we used the monotonicity of E n − , see (3.2). Lemma 3.3.
Let s ∈ A n ( y ) , and t > s be such that u n +2 ( t ) = 2 y . Then E n +1 ( s ) − E n +1 ( t ) > y . Proof.
We have E n +1 ( s ) − E n +1 ( t ) = E n +2 ( s ) − E n +2 ( t ) − u n +2 ( s ) + u n +2 ( t ) > u n +2 ( t ) − u n +2 ( s ) > y . Lemma 3.4.
Let s ∈ A n ( y ) , and assume that t ( s ) = s + 2 λ n +2 + 2 yλ β ( n +2) . (3.5) Then E n ( s ) − E n ( t ( s )) > c y , where c = min (cid:18) λ , λ β (cid:19) Proof.
Formulas (3.4) and (3.5) imply the inequalities u n ( ρ ) > y , u n +2 ( ρ ) y for all ρ ∈ [ s, t ( s )] . So, similarly to (1.1) and using u n +1 ( s ) >
0, we have u n +1 ( τ ) > Z τs λ β ( n +1) u n ( ρ ) exp (cid:18) − Z τρ ( λ n +2 + λ β ( n +2) u n +2 ( σ )) dσ (cid:19) dρ > λ β ( n +1) y Z τs exp (cid:0) − ( λ n +2 + λ β ( n +2) y )( τ − ρ ) (cid:1) dρ = λ β ( n +1) y λ n +2 + 2 yλ β ( n +2) ) (cid:2) − exp (cid:0) − ( λ n +2 + 2 yλ β ( n +2) )( τ − s ) (cid:1)(cid:3) for τ ∈ [ s, t ( s )]. Furthermore, due to (3.1) E n ( s ) − E n ( t ( s )) > Z t ( s ) s (cid:0) λ n u n ( τ ) + λ β ( n +1) u n ( τ ) u n +1 ( τ ) (cid:1) dτ (3.6) > λ n y ( t ( s ) − s )2 + λ β ( n +1) y λ n +2 + 2 yλ β ( n +2) ) Z t ( s ) s h − e − ( λ n +2 +2 yλ β ( n +2) )( τ − s ) i dτ. Let us estimate the last integral. If τ > ( t ( s ) + s ) /
2, then τ − s > t ( s ) − s λ n +2 + 2 yλ β ( n +2) − e − ( λ n +2 +2 yλ β ( n +2) )( τ − s ) > − e − > , and thus, Z t ( s ) s h − e − ( λ n +2 +2 yλ β ( n +2) )( τ − s ) i dτ > t ( s ) − s . Now, (3.6) implies E n ( s ) − E n ( t ( s )) > λ n y ( t ( s ) − s )2 + λ β ( n +1) y ( t ( s ) − s )32( λ n +2 + 2 yλ β ( n +2) )= y (cid:0) λ n +2 + 32 yλ n + β ( n +2) + y λ β ( n +1) (cid:1)
16 ( λ n +2 + 2 yλ β ( n +2) ) > y min (cid:18) λ , λ , λ β (cid:19) . Lemma 3.5.
Let y > , n > M . Let the set A n ( y ) be defined by the formula (3.3) . Then | A n ( y ) | c k a k l y λ βn + y λ n , where c = 4(3 + c )3 c max (cid:18) λ , λ β (cid:19) . Proof.
Let us consider the family I of intervals I = { ( s ; t ( s )) } s ∈ A n ( y ) , where t ( s ) is definedby (3.4). By virtue of Lemma 3.1 there is a finite or countable sequence of disjoint intervals { ( s k ; t k ) } ⊂ I , such that | A n ( y ) | X k ( t k − s k ) . Due to Lemmas 3.2, 3.3 and 3.4, the number of disjoint intervals { ( s k ; t k ) } can not exceed k a k l c y + k a k l y , because the functions E n and E n +1 are non-increasing, see (3.2), and E n (0) E n +1 (0) k a k l .The length of each interval ( t k − s k ) λ n +2 +2 yλ β ( n +2) , by (3.4). So, | A n ( y ) | X k ( t k − s k ) c ) k a k l c y ( λ n +2 + 2 yλ β ( n +2) ) . Lemma 3.6.
Let a n > for n > M , { u n } be a weak solution to (0.1) . For y > we put B n ( y ) = { s > u n ( s ) > y } . Then | B n ( y ) | c k a k l y λ βn + y λ n , for n > M, where c = c (1 − λ − β ) − . roof. As u k ( s ) → k → ∞ , we have B n ( y ) ⊂ ∞ [ k =0 A n +2 k ( y ) . Therefore, | B n ( y ) | P ∞ k =0 | A n +2 k ( y ) | , and the reference to Lemma 3.5 completes the proof. Theorem 3.7.
Let a n > for n > M , and { u n } be a weak solution to (0.1) . Then Z ∞ u n ( t ) dt c k a k l λ − βn log (cid:0) λ ( β − n k a k l + 1 (cid:1) , where the constant c is defined in Lemma 3.6.Proof. By Lemma 1.2 we have B n ( y ) = ∅ if y > k a k l . Therefore, due to Lemma 3.6 Z ∞ u n ( t ) dt = Z k a k l | B n ( y ) | y dy c k a k l Z k a k l dyλ βn y + λ n = 3 c k a k l λ − βn log (cid:0) λ ( β − n k a k l + 1 (cid:1) . l +2 Lemma 4.1.
Let a n > for n > M . Let { u n } and { v n } be two weak solutions to (0.1) . If weset ψ N ( t ) = N X n =1 − n ( u n ( t ) − v n ( t )) , then for N > M we have ˙ ψ N ( t ) λ β ( M +1) k a k l ψ N ( t ) + λ β ( N +1) − N (cid:0) u N ( t ) v N +1 ( t ) + u N +1 ( t ) v N ( t ) (cid:1) . Proof.
Put w n ( t ) = u n ( t ) − v n ( t ). We have12 N X n =1 ddt (cid:18) w n n (cid:19) + N X n =1 λ n n w n = N X n =1 w n n (cid:0) λ βn ( u n − − v n − ) − λ β ( n +1) ( u n u n +1 − v n v n +1 ) (cid:1) = N X n =1 (cid:18) λ βn n w n − w n ( u n − + v n − ) − λ β ( n +1) n +1 w n ( w n ( u n +1 + v n +1 ) + w n +1 ( u n + v n )) (cid:19) = − N X n =1 λ β ( n +1) n +1 w n ( u n +1 + v n +1 ) − λ β ( N +1) N +1 w N w N +1 ( u N + v N ) − M X n =1 λ β ( n +1) n +1 w n ( u n +1 + v n +1 ) − λ β ( N +1) N +1 ( u N − v N )( u N +1 − v N +1 ) . In the last sum we have changed the upper limit N by M , because u n , v n > n > M . Dueto Lemma 1.2 | u n ( t ) | , | v n ( t ) | k a k l , therefore12 ˙ ψ N ( t ) λ β ( M +1) k a k l ψ M ( t ) + λ β ( N +1) − N − (cid:0) u N ( t ) v N +1 ( t ) + u N +1 ( t ) v N ( t ) (cid:1) . heorem 4.2. Let { a n } ∈ l +2 , where the set l +2 is defined by (1.2) . Then the weak solution to (0.1) is unique.Proof. Let { u n } and { v n } be weak solutions to (0.1). Put ψ N ( t ) = N X n =1 − n ( u n ( t ) − v n ( t )) , ψ ∞ ( t ) = ∞ X n =1 − n ( u n ( t ) − v n ( t )) . Note, that the sequence ψ n tends to ψ ∞ uniformly on [0 , ∞ ) due to the fact that all the functions { u n } , { v n } are uniformly bounded by k a k l .We have a n > n > M for some M . Lemma 4.1 implies that for N > Mψ N ( t ) λ β ( M +1) k a k l Z t ψ N ( τ ) dτ (4.1)+2 − N λ β ( N +1) Z t (cid:0) u N ( τ ) + u N +1 ( τ ) + v N ( τ ) + v N +1 ( τ ) (cid:1) dτ. Due to Theorem 3.7 the last integral in (4.1) is O ( N λ − βN ). Passing in (4.1) to the limit N → ∞ , we obtain ψ ∞ ( t ) λ β ( M +1) k a k l Z t ψ ∞ ( τ ) dτ. (4.2)This inequality yields ψ ∞ ≡
0, and u n ≡ v n .In this proof we followed almost literally [2], where such uniqueness is established for thesystem (0.3). The difference is that in [2] all a n are supposed to be non-negative, whereas wesuppose that a n > M . It leads to the appearance of the first term inthe right hand side of (4.1), and of the non-trivial right hand side in (4.2). Nevertheless, (4.2)implies that the function ψ ∞ is identically zero. Proof of Theorem 0.7.
The existence of a Leray-Hopf solution is proved in Theorem 2.2. Let usestablish its uniqueness. Let { u n } be a Galerkin solution, and { v n } be a Leray-Hopf solutionto (0.1), see Definitions 2.1 and 0.2. It is sufficient to prove that u n ≡ v n . Let ε < ε , where ε is defined in (0.5), and assume that | a n | ε λ (2 − β ) n for n > K. Let us define t as t = ( inf { t : { u n ( t ) } ∈ l +2 } , if this set is non-empty , + ∞ , if { u n ( t ) } / ∈ l +2 for all t ;the set l +2 is defined in (1.2). Assume first that 0 < t < ∞ .There are two possibilities.1) Let { u n ( t ) } ∈ l +2 . For any t < t we have { u n ( t ) } / ∈ l +2 , and by virtue of Corollary 2.8there is a number L such that | u n ( t ) | ε λ (2 − β ) n for all n > L, t ∈ [0 , t ] . u n ≡ v n on [0 , t ] due to Theorem 1.4. It is true for all t < t , and thus u n ≡ v n on[0 , t ], as all these functions are continuous. In particular, u n ( t ) = v n ( t ), and { u n ( t ) } ∈ l +2 .Now, we can apply Theorem 4.2 on the interval [ t , ∞ ), and therefore u n ≡ v n on [ t , ∞ ) aswell.2) Let { u n ( t ) } / ∈ l +2 . In this case we can apply the Corollary 2.8 on [0 , t ], which meansthat there is a number L such that | u n ( t ) | ε λ (2 − β ) n for all n > L, t ∈ [0 , t ] . In particular, | u n ( t ) | ε λ (2 − β ) n , n > L . By Corollary 2.7 there exists a time t > t suchthat there is a Galerkin solution { ˜ u n ( t ) } to the problem (0.1) on the interval [ t , t ] with theinitial data ˜ u n ( t ) = u n ( t ), and moreover, | ˜ u n ( t ) | ελ (2 − β ) n for n > L, t ∈ [ t , t ] . Furthermore, { ˜ u n } is a Leray-Hopf solution on [ t , t ], because any Galerkin solution is a Leray-Hopf solution, see subsection 2.1. We putˆ u n ( t ) = ( u n ( t ) , t t , ˜ u n ( t ) , t < t t . This { ˆ u n } is a Leray-Hopf solution to (0.1) on the interval [0 , t ], and | ˆ u n ( t ) | ελ (2 − β ) n for n > L, t ∈ [0 , t ] . By Theorem 1.4 { ˆ u n } coincides with { u n } and with { v n } , ˆ u n ≡ u n ≡ v n , on [0 , t ]. By definitionof the time t we have { u n ( t ) } ∈ l +2 , see Lemma 1.1. Now, u n ≡ v n also on [ t , ∞ ) due toTheorem 4.2.The cases t = 0 and t = ∞ can be treated in the same way. In this section we study the equation (0.6) and prove Theorem 0.10. There are no differentialequations here, the theory of number sequences only. β < Following [3] we change the variables a n = − λ (2 − β ) n − b n . (5.1)Then (0.6) is equivalent to the system b n b n +1 = b n + ub n − , b = 0 , n = 1 , , . . . , (5.2)where u = λ β − . In this subsection we show that if λ β − < β <
3) then there exists a sequence of positive numbers { b n } satisfying (5.2). In order to satisfy(5.2) for n = 1 we take b = 1. 16ne can consider (5.2) as a recurrent formula which allows to express b n +1 via b n − and b n .It turns out that it is more convenient to make the process run backward, and find b n − via b n and b n +1 , b n − = p b n ( b n +1 − /u . We construct an auxiliary number sequence { c n } , which isdefined by two first terms c , c >
1, and by the following rule:if c k − > , then c k +1 = r c k ( c k − − u ; (5.3)if c k −
1, then the sequence stops at c k .Intruduce the notations κ = 12 + r
14 + 1 u , ν = 14 r u ! . (5.4)It is easy to see that κ > ν > u < Lemma 5.1.
Let δ > be such that κ δ < u − u . Assume that c > − u − δ, c > − u − κ δ (clearly, c , c > ). Then c > − u − κ δ. Proof.
We have c = 1 √ u p c ( c − > − u s (1 − κ δ (1 − u )) (cid:18) − δu (1 − u ) (cid:19) > − u (1 − κ δ (1 − u )) (cid:18) − δu (1 − u ) (cid:19) > − u − κ δ − δu = 11 − u − κ δ by definition of κ , see (5.4).By induction we obtain Corollary 5.2.
Assume that κ n δ < u − u , c > − u − δ, c > − u − κ δ. Then the sequence { c k } is defined at least until k = n + 1 , and c n > − u − κ n δ . Lemma 5.3.
Let δ > be such that νδ < u − u , < c − u − δ, < c − u − νδ. Then c − u − ν δ. roof. We have c − u s (1 − νδ (1 − u )) (cid:18) − δu (1 − u ) (cid:19) − u ) (cid:18) − νδ (1 − u ) + 1 − δu (1 − u ) (cid:19) = 11 − u − (cid:18) ν + 1 u (cid:19) δ = 11 − u − ν δ by definition of ν , see (5.4). Corollary 5.4.
Let ν m δ < u − u , c − u − δ , c − u − νδ . Assume that the sequence { c k } is defined until k = m , i.e. c k > for k = 0 , , . . . , m − . Then c m − u − ν m δ. Lemma 5.5.
Let M ∈ N . There exists a positive number δ such that if c = 11 − u − δ , c = 11 − u − νδ , where ν is defined in (5.4) , then the sequence { c k } is defined until k = n + 1 with some n > M ,and c k < − u ∀ k, and c n < . Proof.
Choose δ such that κ M δ < u − u . Then by Corollary 5.2 c , c , . . . , c M >
1. On theother hand, Corollary 5.4 implies that the sequence { c k } can not be infinite. So, there is n > M such that c n <
1. The inequality c k < − u for all k follows also from Corollary 5.4. Corollary 5.6.
Let M ∈ N . There exists a positive number δ such that if c = 11 − u − δ , c = 11 − u − νδ , then the sequence { c k } is defined until k = n + 1 with some n > M , and c k < − u ∀ k, and c n = 1 . Proof.
Put c = − u − δ , c = − u − νδ with δ ∈ [0 , δ ]. We consider the elements of thesequence { c k } n +1 k =0 as the functions of δ ∈ [0 , δ ]. Here δ and n are the numbers from Lemma5.5. Clearly, all the functions c k ( δ ) are continuous and decreasing in δ . Note, that c k (0) = − u for all k , and c n ( δ ) < < − u = c n (0) . Therefore, there exists δ such that c n ( δ ) = 1. Corollary 5.7.
Let M ∈ N . There is a number n > M , and a positive number d , such thatthe sequence { d k } , defined by two first terms { d , d } with d = 1 , and by the recurrent formula d k +1 = 1 + ud k − d k , (5.5) obeys the property d k < − u for k n + 2 . roof. It is sufficient to take d k = c n +2 − k , k = 1 , . . . , n + 2 , where { c k } is the sequence from the precedent Corollary. Theorem 5.8.
Let u < . Then there is an infinite number sequence { b k } , b k > , b = 1 ,satisfying (5.2) and such that b k − u for all k .Proof. By virtue of Corollary 5.7 there are a sequence of numbers { n m } , n m → ∞ , and asequence of sequences { d ( m ) k } k =1 ,...,n m , such that d ( m )2 = 1 , d ( m ) k +1 = 1 + u (cid:16) d ( m ) k − (cid:17) d ( m ) k , d ( m ) k < − u . The sequence of the first elements { d ( m )1 } ∞ m =1 is bounded. Without loss of generality one canassume that this sequence converges, d ( m )1 −→ m →∞ b . Each element d ( m ) k is a continuous functionof d ( m )1 . Therefore, d ( m ) k −→ m →∞ b k , and it is clear that b k +1 = 1 + ub k − b k , and b k − u for all k. Taking into account the change (5.1) we see that Theorem 5.8 means the existence of non-trivial solution { a n } to the system (0.6) such that | a n | λ (2 − β ) n (1 − u ) λ . The estimate | a n | > λ (2 − β ) n − for any non-trivial solution is done in Theorem 0.9. Thus, thecase β < Remark 5.9.
The sequence { b k } from Theorem 5.8 converges, b k → − u . Indeed, the sequence { c k } from Corollary 5.6 decreases (the inequality c > c > c implies c > c , and one canproceed by induction). So, the sequence { b k } increases, and therefore converges, b k → b ∞ .Finally, (5.2) yields b ∞ = − u . β = 3 In this subsection we study the equation (5.2) with u = 1. Lemma 5.10.
Let b , b > , b n +1 = 1 + b n − b − n . Then the sequence { b n } is unbounded.Proof. Put lim sup k →∞ b k +1 = K, lim sup k →∞ b k = L. Suppose that
K, L < ∞ . Without loss of generality one can assume K > L . Let us pick ε > K − ε ) L + ε > K + ε.
19y definition of K and L , we can choose a number N such that b k L + ε, b k +1 K + ε, for all k > N, and a number M > N such that b M − > K − ε . Then we have b M +1 = 1 + b M − b M > K − ε ) L + ε > K + ε, which is a contradiction.Now we consider the sequence { a n } , see (5.1). Several first terms can vanish, but if there isa non-zero term in the sequence, then the Lemma 5.10 means thatlim sup n →∞ a n λ n = + ∞ for β = 3 . In the rest of this subsection we show that there is a sequence satisfying (5.2) with u = 1,which increases like a linear function. Again, we consider an auxiliary sequence { c n } , definedby the rule: if c k − > , then c k +1 = p c k ( c k − − , if c k − , then the sequence stops at c k . Lemma 5.11. If c A , c A − / , then c k A − k/ .Proof. Clearly, c p ( A − / A − A − /
3. Further we proceed by induction.
Lemma 5.12.
Let c = A , c = A − / . Then the inequalities c k +1 < c k < c k +1 + 1 , (5.6) c k > A − k, (5.7) are fulfilled.Proof. As c A − / < c , we obtain by induction again c k +1 < c k . On the other hand, thisinequality means c k +1 = p c k ( c k − − < c k ⇒ c k − − < c k , and (5.6) is proved. The estimate (5.7) follows from (5.6). Lemma 5.13.
Let M ∈ N , A > M + 1 . If c = A , c = A − / , then the sequence { c k } isdefined until k = n + 1 , n > M , and c n < .Proof. Due to (5.7) c M > A − M >
1, where from n > M . By virtue of Lemma 5.11, thereexists n such that c n < Corollary 5.14.
Let M ∈ N . There is A such that if c = A , c = A − / , then thesequence { c k } is defined until k = n + 1 , n > M , and c n = 1 . orollary 5.15. Let M ∈ N . There is n > M and d ∈ (0 , such that the sequence { d k } defined by two first elements { d , d } , d = 1 , and by the recurrent formula d k +1 = 1 + d k − d − k satisfies the condition d k < k for k n . In the proof of this Corollary one uses the inequality (5.6).
Theorem 5.16.
There exists an infinite sequence { b k } , b k > , b = 1 , such that b k +1 = 1 + b k − b k and b k k for all k. Thus, Theorem 0.10 in the part β = 3 is proved. β > Let us do the change a n = − e n λ − βn/ in (0.6). Then (0.6) is equivalent to the system e n e n +1 = λ ((6 − β ) n − β ) / e n + e n − . By Theorem 0.9 e n >
0, therefore e n − max( e n , e n +1 ). Iterating this inequality we get e n − max( e n , e n +1 ) max( e n +1 , e n +2 ) . . . max( e m , e m +1 )for all m > n . So, lim sup n →∞ e n > e k = 0. We proved Lemma 5.17.
Let β > , { a n } be a non-trivial solution to (0.6) . Then lim sup n →∞ ( − a n ) λ βn > . Let us show that decreasing like λ − βn/ is possible. We return to the equation (5.2), andconsider the auxiliary sequence (5.3); now u > Lemma 5.18.
Let c = A , c = Au − / . Then we have the following relations:a) c < Au − / ;b) c k +1 < c k u − / ;c) c k − < c k u / + 1 ;d) c k > ( A − u − k/ − ( u / − − .Proof. a) Clearly, c = p c ( c − u − = q A ( A − u − / < Au − / . b) follows from a) by induction.c) We have c k +1 = p c k ( c k − − u − < c k u − / due to the point b). Therefore, c k − − < c k u / .d) follows from c) by induction. Corollary 5.19.
Let M ∈ N . There exists a number A such that if c = A , c = A u − / ,then the sequence { c k } is defined until k = n + 1 , n > M , and c n = 1 . Lemma 5.20.
Let M ∈ N . There exist a number n > M and a number d ∈ (0 , such thatthe sequence { d k } defined by two first elements { d , d } , d = 1 , and by the formula (5.5) with u > , satisfies the condition d k +1 u k/ − u / − for k n + 2 . (5.8) Proof.
We put d k = c n +2 − k , where { c k } is the sequence from the Corollary 5.19. Due to Lemma5.18 c) we have d k +1 < d k u / + 1. Taking into account that d = 1, we get from here (5.8).In the same manner as in the proof of the Theorem 5.8, we get from here Theorem 5.21.
Let u > . There exists an infinite sequence { b k } ∞ k =1 , b k > , b = 1 , satisfying (5.2) and such that b k +1 u k/ − u / − for all k. By virtue of (5.1) this Theorem means that there is a non-trivial solution { a n } to the system(0.6) such that a n = O ( λ − βn/ ), n → ∞ . Thus, the part β > Remark 5.22.
In [1] the Theorem 5.21 is proved, and moreover, the existence of a positivelimit lim n →∞ b n u − n/ is shown. We provided our proof for the sake of completeness, as it is verysimilar to the cases β < β = 3. In [1], the equation (5.2) appears in studying of self-similarsolution to the problem (0.3). Let { b n } be a solution to (5.2) with u = κ >
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