Existence, uniqueness and global behavior of the solutions to some nonlinear vector equations in a finite dimensional Hilbert space
aa r X i v : . [ m a t h . C A ] D ec Existence, uniqueness and global behavior of thesolutions to some nonlinear vector equations in afinite dimensional Hilbert space
Mama ABDELLILaboratory of Analysis and Control of Partial Differential Equations,Djillali Liabes University, P. O. Box 89, Sidi Bel Abbes 22000, Algeria.abdelli − [email protected]´ıa ANGUIANODepartamento de An´alisis Matem´atico, Facultad de Matem´aticas,Universidad de Sevilla, P. O. Box 1160, 41080-Sevilla, [email protected] HARAUXSorbonne Universit´es, UPMC Univ Paris 06, CNRS, UMR 7598,Laboratoire Jacques-Louis Lions, 4, place Jussieu 75005, Paris, [email protected] Abstract
The initial value problem and global properties of solutions are studied for thevector equation: (cid:16) k u ′ k l u ′ (cid:17) ′ + k A u k β Au + g ( u ′ ) = 0 in a finite dimensional Hilbertspace under suitable assumptions on g . AMS classification numbers:
Keywords:
Existence of the solution, Uniqueness of the solution, Decay rate.1
Introduction
Let H be a finite dimensional real Hilbert space, with norm denoted by k . k . We considerthe following nonlinear equation (cid:16) k u ′ k l u ′ (cid:17) ′ + k A u k β Au + g ( u ′ ) = 0 , (1.1)where l and β are positive constants, and A is a positive and symmetric linear operatoron H . We denote by ( · , · ) the inner product in H . The operator A is coercive, whichmeans : ∃ λ > , ∀ u ∈ D ( A ) , ( Au, u ) ≥ λ k u k . We also define ∀ u ∈ H, k A u k := k u k D ( A ) , a norm equivalent to the norm in H . We assume that g : H → H is locally Lipschitzcontinuous.When l = 0 and g ( u ′ ) = c | u ′ | α u ′ Haraux [5] studied the rate of decay of the energyof non-trivial solutions to the scalar second order ODE. In addition, he showed that if α > ββ +2 all non-trivial solutions are oscillatory and if α < ββ +2 they are non-oscillatory.In the oscillatory case he established that all non-trivial solutions have the same decayrates, while in the non-oscillatory case he showed the coexistence of exactly two differentdecay rates, calling slow solutions those which have the lowest decay rate, and fastsolutions the others.Abdelli and Haraux [1] studied the scalar second order ODE where g ( u ′ ) = c | u ′ | α u ′ ,they proved the existence and uniqueness of a global solution with initial data ( u , u ) ∈ R . They used some modified energy function to estimate the rate of decay and theyused the method introduced by Haraux [5] to study the oscillatory or non-oscillatoryof non-trivial solutions. If α > β ( l +1)+ lβ +2 all non-trivial solutions are oscillatory and if α < β ( l +1)+ lβ +2 they are non-oscillatory. In the non-oscillatory rate, as in the case l = 0,the coexistence of exactly two different decay rates was established.In this article, we use some techniques from Abdelli and Haraux [1] to establisha global existence and uniqueness result of the solutions, and under some additionalconditions on g (typically g ( s ) ∼ c k s k α s ), we study the asymptotic behavior as t → ∞ .A basic role will be played by the total energy of the solution u given by the formula E ( t ) = l + 1 l + 2 k u ′ ( t ) k l +2 + 1 β + 2 k A u ( t ) k β +2 . (1.2)The plan of this paper is as follows: In Section 2 we establish some basic preliminaryinequalities, and in Section 3 we prove the existence of a solution u ∈ C ( R + , H ) with k u ′ k l u ′ ∈ C ( R + , H ) for any initial data ( u , u ) ∈ H × H under relevant conditions on g and the conservation of total energy for each such solution. In Section 4 we establish theuniqueness of the solution in the same regularity class under additional conditions on g . In Section 5 we prove convergence of all solutions to 0 under more specific conditionson g and we estimate the decay rate of the energy. Finally, in Section 6, we discuss the2ptimality of these estimates when g ( s ) = c k s k α s and l < α < β (1+ l )+ lβ +2 ; in particular,by relying on a technique introduced by Ghisi, Gobbino and Haraux [3] , we prove theexistence of a open set of initial states giving rise to slow decaying solutions. In ourlast result, by relying on a technique introduced by Ghisi, Gobbino and Haraux [4], weprove that all non-zero solutions are either slow solutions or fast solutions. In this section, we establish some easy but powerful lemmas which generalize Lemma2.2 and Lemma 2.3 from [2] and will be essential for the existence and uniqueness proofsof the next section. Troughout this section, H denotes an arbitrary (not necessary finitedimensional) real Hilbert space with norm denoted by k . k . Proposition 2.1.
Let ( u, v ) ∈ H × H and ( α, β ) ∈ IR + × IR + . Then the condition ( α − β )( k u k − k v k ) ≥ , implies ( αu − βv )( u − v ) ≥
12 ( α + β ) k u − v k . Proof.
An easy calculation gives the identity( αu − βv )( u − v ) = 12 ( α − β )( k u k − k v k ) + 12 ( α + β ) k u − v k . Since ( α − β )( k u k − k v k ) = ( k u k + k v k )( α − β )( k u k − k v k ) , the result follows immediately.For the next results, we consider a number R > J R = [0 , R ]; B R := { u ∈ H, k u k ≤ R } Proposition 2.2.
Let f : J R −→ IR + be non increasing and such that for some positivenumbers p , c : ∀ s ∈ J R , f ( s ) ≥ cs p . Then we have ∀ ( u, v ) ∈ B R × B R , ( f ( k u k ) u − f ( k v k ) v )( u − v ) ≥ c k u k p + k v k p ) k u − v k . Proof.
Applying the previous result with α = f ( k u k ) and β = f ( k v k ) the result followsimmediately. 3 orollary 2.3. Let f : J R −→ IR + be non increasing and such that for some positivenumbers p , c : ∀ s ∈ J R , f ( s ) ≥ cs p . Then we have ∀ ( u, v ) ∈ B R × B R , k f ( k u k ) u − f ( k v k ) v ) k ≥ c k u k p + k v k p ) k u − v k . Proof.
This inequality follows immediately from Cauchy-Schwarz inequality combinedwith the conclusion of the previous proposition.
Proposition 2.4.
Let f : J R −→ IR + be non increasing and such that for some positivenumbers p , c : ∀ s ∈ J R , f ( s ) ≥ cs p . Then we have ∀ ( u, v ) ∈ B R × B R , ( f ( k u k ) u − f ( k v k ) v )( u − v ) ≥ δ k u − v k p +2 , with δ = c max { p, } .Proof. It is sufficient to apply the previous result combined with the inequality k u − v k p ≤ max { p − , } ( k u k p + k v k p ) . Corollary 2.5.
Let f : J R −→ IR + be non increasing and such that for some positivenumbers p , c : ∀ s ∈ J R , f ( s ) ≥ cs p . Then we have for some constant C = C ( c, p ) > , ∀ ( u, v ) ∈ B R × B R , k u − v k ≤ C k f ( k u k ) u − f ( k v k ) v ) k p +1 . Proof.
This inequality follows immediately from Cauchy-Schwarz inequality combinedwith the conclusion of the previous proposition.
Lemma 2.6.
Assume that A is a positive, symmetric, bounded operator on H . forsome constant D > we have ∀ ( w, v ) ∈ H × H ∀ ( w, v ) ∈ H × H, kk A w k β Aw − k A v k β Av k ≤ D max( k A v k , k A w k ) β k w − v k . Proof.
We can write kk A w k β Aw − k A v k β Av k = k A ( k A w k β A w − k A v k β A v ) k≤ k A kkk A w k β A w − k A v k β A v k . Direct calculations (see also [2] , Lemma 2.2) yield kk A w k β A w − k A v k β A v k ≤ C max( k A v k , k A w k ) β k A w − A v k≤ C max( k A v k , k A w k ) β k w − v k , and the result is proved. 4 Global existence and energy conservation for equation (1.1)
In this section, we study the existence of a solution for the initial value problem associ-ated to equation (1.1) where g : H → H is a locally Lipschitz continuous function whichsatisfies the following hypothesis: ∃ k > , k > , ∀ v, ( g ( v ) , v ) ≥ − k − k k v k l +2 . (3.1) Theorem 3.1.
Let ( u , u ) ∈ H × H . The problem (1.1) has a global solution satisfying u ∈ C ( R + , H ) , k u ′ k l u ′ ∈ C ( R + , H ) and u (0) = u , u ′ (0) = u . Proof.
To show the existence of the solution for (1.1), we consider the auxiliary problem (cid:26) ( ε + k u ′ ε k ) l/ u ′′ ε + l ( u ′ ε , u ′′ ε )( ε + k u ′ ε k ) l/ − u ′ ε + k A u ε k β Au ε + g ( u ′ ε ) = 0 ,u ε (0) = u , u ′ ε (0) = u . (3.2) Here, ε > l/ m > . Assuming the existence of such a solution u ε , multiplying (3.2) by u ′ ε , we find[( ε + k u ′ ε k ) m + 2 m ( ε + k u ′ ε k ) m − k u ′ ε k ]( u ′ ε , u ′′ ε ) + k A u ε k β ( Au ε , u ′ ε ) + ( g ( u ′ ε ) , u ′ ε ) = 0 , then ( u ′ ε , u ′′ ε ) = − k A u ε k β ( Au ε , u ′ ε ) + ( g ( u ′ ε ) , u ′ ε )( ε + k u ′ ε k ) m − ( ε + ( l + 1) k u ′ ε k ) . (3.3)From (3.2), we obtain that u ε is a solution of ( ε + k u ′ ε k ) m u ′′ ε − l k A u ε k β ( Au ε , u ′ ε ) + ( g ( u ′ ε ) , u ′ ε ) ε + ( l + 1) k u ′ ε k u ′ ε + k A u ε k β Au ε + g ( u ′ ε ) = 0 ,u ε (0) = u , u ′ ε (0) = u . (3.4) Conversely, (3.4) implies (3.3). Then, replacing (3.3) in (3.4), we obtain (3.2). Therefore(3.2) is equivalent to (3.4).i) A priori estimates:Next, we introduce F ε ( u ε , u ′ ε ) = l k A u ε k β ( Au ε , u ′ ε ) + ( g ( u ′ ε ) , u ′ ε )( ε + k u ′ ε k ) m ( ε + ( l + 1) k u ′ ε k ) u ′ ε − k A u ε k β Au ε + g ( u ′ ε )( ε + k u ′ ε k ) m . Then (3.4), can be rewritten as (cid:26) u ′′ ε = F ε ( u ε , u ′ ε ) ,u ε (0) = u , u ′ ε (0) = u . (3.5) F ε is locally Lipschitz continuous, the existence and unique-ness of u ε in the class C ([0 , T ) , H ), for some T > u ′ ε , we obtain by a simple calculation the following energy identity: ddt E ε ( t ) + ( g ( u ′ ε ( t )) , u ′ ε ( t )) = 0 , where E ε ( t ) = l + 1 l + 2 ( ε + k u ′ ε k ) m +1 − ε ( ε + k u ′ ε k ) m + 1 β + 2 k A u ε ( t ) k β +2 . Indeed, for any function v ∈ C ([0 , T ) , H ) we have the following sequence of iden-tities ((( ε + k v k ) m v ) ′ , v ) = 2 m ( ε + k v k ) m − k v k ( v, v ′ ) + ( ε + k v k ) m ( v, v ′ )= 2 m ( ε + k v k ) m − ( ε + k v k )( v, v ′ ) − mε ( ε + k v k ) m − ( v, v ′ ) + ( ε + k v k ) m ( v, v ′ )= (2 m + 1)( ε + k v k ) m ( v, v ′ ) − mε ( ε + k v k ) m − ( v, v ′ )= ddt (cid:20) l + 1 l + 2 ( ε + k v k ) m +1 − ε ( ε + k v k ) m (cid:21) . Moreover for some constant
C > ε , we have E ε ( t ) + C ≥ l + 1 l + 2 ( ε + k u ′ ε k ) m +1 − ε ( ε + k u ′ ε k ) m + C ≥ k u ′ ε ( t ) k l +2 and then as a consequence of (3.1) ddt E ε ( t ) = − ( g ( u ′ ε ( t )) , u ′ ε ( t )) ≤ k + k k u ′ ε ( t ) k l +2 ≤ k + k E ε ( t ) . By Gronwall’s inequality, this implies ∀ t ∈ [0 , T ) , k u ε ( t ) k ≤ M , k u ′ ε ( t ) k ≤ M , (3.6)for some constants M , M independent of ε . Hence, u ε and u ′ ε are uniformlybounded and u ε is a global solution, in particular T > ε → u ε and u ′ ε areuniformly equicontinous on (0 , T ) for any T >
0. For u ε it is clear, since k u ′ ε k isbounded.Moreover we have( ε + k u ′ ε k ) l/ u ′′ ε + l ( u ′ ε , u ′′ ε )( ε + k u ′ ε k ) l/ − u ′ ε = (( ε + k u ′ ε k ) l/ u ′ ε ) ′ , hence by the equation k (( ε + k u ′ ε k ) l/ u ′ ε ) ′ k ≤ k A u ε ( t ) k β k Au ε ( t ) k + k g ( u ′ ε ( t )) k , g is locally Lipschitz continuous, hence bounded on bounded sets, weobtain k (( ε + k u ′ ε k ) l/ u ′ ε ) ′ k ≤ M . Therefore the functions ( ε + k u ′ ε k ) m u ′ ε are uniformly Lipschitz continuous on R +.We claim that u ′ ε is a uniformly (with respect to ε ) locally H¨older continuousfunction of t . Indeed by applying Corollary 2.5 with f ( s ) = ( ε + s ) m and p =2 m = l we find for some C > k u ′ ε ( t ) − u ′ ε ( t ) k ≤ C k ( ε + k u ′ ε k ) m u ′ ε ( t ) − ( ε + k u ′ ε k ) m u ′ ε ( t ) k l +1 ≤ C ′ k t − t k l +1 . As a consequence of Ascoli’s theorem and a priori estimate (3.6) combined with(3.5), we may extract a subsequence which is still denoted for simplicity by ( u ε )such that for every T > u ε → u in C ((0 , T ) , H ) , as ε tends to 0. Integrating (3.2) over (0 , t ), we get( ε + k u ′ ε k ) m u ′ ε ( t ) − ( ε + k u ′ ε k ) m u ′ ε ( t )(0) (3.7)= − Z t k A u ε ( s ) k β Au ε ( s ) ds − Z t g ( u ′ ε ( s )) ds. From (3.7), we then have, as ε tends to 0( ε + k u ′ ε k ) m u ′ ε ( t ) → − Z t k A u ( s ) k β Au ( s ) ds − Z t g ( u ′ ( s )) ds + k u ′ (0) k l u ′ (0) in C ((0 , T ) , H ) . Hence k u ′ k l u ′ = − Z t k A u ( s ) k β Au ( s ) ds − Z t g ( u ′ ( s )) ds + k u ′ (0) k l u ′ (0) , (3.8)and k u ′ k l u ′ ∈ C ((0 , T ) , H ). Finally by differentiating (3.8) we conclude that u isa solution of (1.1). Remark 3.2.
It is not difficult to see that the solution u constructed in the existencetheorem satisfies the energy identity ddt E ( t ) = − ( g ( u ′ ( t )) , u ′ ( t )) . The following strongerresult shows that this identity is true for any solution even if uniqueness is not known.For infinite dimensional equations such as the Kirchhoff equation, both uniqueness andthe energy identity for general weak solutions are old open problems. Theorem 3.3.
Let ( u , u ) ∈ H × H . Then any solution u of (1.1) such that u ∈ C ( R + , H ) , k u ′ k l u ′ ∈ C ( R + , H ) and u (0) = u , u ′ (0) = u , satisfies the formula ddt E ( t ) = − ( g ( u ′ ( t )) , u ′ ( t )) . (3.9)7he proof of this new result relies on the following simple lemma Lemma 3.4.
Let J be any interval, assume that v ∈ C ( J, H ) and k v k l v ∈ C ( J, H ) then k v ( t ) k l +2 ∈ C ( J, H ) , and (( k v ( t ) k l v ( t )) ′ , v ( t )) = ddt (cid:16) l + 1 l + 2 k v ( t ) k l +2 (cid:17) , ∀ t ∈ J. Proof.
Case 1.
For any t ∈ J , if v ( t ) = 0 then v ( t ) = 0 in the neighborhood of t and in this neighborhood we have v ∈ C ( J, H ) with ddt (cid:0) k v ( t ) k l +2 (cid:1) = ddt (cid:0) k v ( t ) k (cid:1) l/ = (cid:18) l (cid:19) ( k v ( t ) k ) l/ ddt (cid:0) k v ( t ) k (cid:1) = 2 (cid:18) l (cid:19) k v ( t ) k l ( v ( t ) , v ′ ( t )) . (3.10) On the other hand, we find (( k v ( t ) k l v ( t )) ′ , v ( t )) = ( k v ( t ) k l v ′ ( t ) , v ( t )) + ( k v ( t ) k l ) ′ ( v ( t ) , v ( t ))= k v ( t ) k l ( v ′ ( t ) , v ( t )) + l k v ( t ) k l ( v ′ ( t ) , v ( t ))= ( l + 1) k v ( t ) k l ( v ′ ( t ) , v ( t )) , (3.11) and from (3.10)-(3.11), we obtain(( k v ( t ) k l v ( t )) ′ , v ( t )) = ddt h l + 1 l + 2 k v ( t ) k l +2 i , for t = t . Case 2. If v ( t ) = 0 then since k v ( t ) k l v ( t ) ∈ C ( J, H ) , we have clearly(( k v k l v ) ′ ( t ) , v ( t )) = 0 . Moreover in the neighborhood of t we have k v ( t ) k l v ( t ) = 0( | t − t | ) , in other terms k v ( t ) k l +1 ≤ C | t − t | , therefore k v ( t ) k l +2 ≤ C l +2 l +1 | t − t | l +1 , then ( k v ( t ) k l +2 ) ′ = 0 , for t = t , and finally (( k v ( t ) k l v ( t )) ′ , v ( t )) = ddt h l + 1 l + 2 k v ( t ) k l +2 i = 0 for t = t .
8e now give the proof of Theorem 3.3.
Proof.
Setting v = u ′ , from Lemma 3.4, we deduce(( k u ′ ( t ) k l u ′ ( t )) ′ , u ′ ( t )) = ddt h l + 1 l + 2 k u ′ ( t ) k l +2 i , ∀ t ∈ J. By multiplying equation (1.1) by u ′ , we obtain easily ddt E ( t ) = − ( g ( u ′ ( t )) , u ′ ( t )) . ( u , u ) given In this section we suppose that ∀ R ∈ IR + , ∀ ( u, v ) ∈ B R × B R , k g ( u ) − g ( v ) k ≤ k ( R ) k u − v k , (4.1)for some positive constant k ( R ). Theorem 4.1.
Let ( u , u ) ∈ H × H , J an interval of R and t ∈ J . Then (1.1) hasat most one solution u ∈ C ( J, H ) , k u ′ k l u ′ ∈ C ( J, H ) and u ( t ) = u , u ′ ( t ) = u . Remark 4.2.
The uniqueness of solutions of (1.1) will be proved under conditions onthe initial data ( u , u ) . The next proposition concerns the uniqueness result for u = 0 . Proposition 4.3.
Let τ ∈ R + and J = ( τ, T ) , T > τ . Then there is at most onesolution of (1.1) with u ( τ ) = u and u ′ ( τ ) = u for T − τ small enough such that u ∈ C ( J, H ) , and k u ′ k l u ′ ∈ C ( J, H ) . Proof.
Since u ′ ( τ ) = 0, the second derivative u ′′ ( τ ) exists and u ′′ ( t ) also exists for τ ≤ t < τ + ε with ε small enough.On ( T, T + ε ), (1.1) reduces to u ′′ = l k u ′ k l − u ′ k A u k β ( Au, u ′ ) + ( g ( u ′ ) , u ′ )( l + 1) k u ′ k l k u ′ k l − k A u k β Au + g ( u ′ ) k u ′ k l , the existence and uniqueness of u in the class C ( J, H ) for this equation is classical.
Proposition 4.4.
Let a = 0 . Then for J an interval containing and such that | J | issmall enough, equation (1.1) has at most one solution satisfying u ∈ C ( J, H ) , k u ′ k l u ′ ∈ C ( J, H ) and u = a, u = 0 . roof. From (1.1), we obtain( k u ′ k l u ′ ) ′ (0) = −k A a k β Aa = ξ, where k ξ k 6 = 0.We set k u ′ k l u ′ = ψ ( t ), then ψ (0) = 0. For any t = 0 we have ψ ( t ) t = 1 t Z t ψ ′ ( s ) ds. It follows that ψ ( t ) t → ξ as t →
0, therefore for | t | small enough we have, k ψ ( t ) k ≥ t k ξ k . Hence, k u ′ k l ≥ ηt ll +1 for | t | small enough and some η > , t ), we have since u ′ (0) = 0 k u ′ ( t ) k l u ′ ( t ) = − Z t k A u ( τ ) k β Au ( τ ) dτ − Z t g ( u ′ ( τ )) dτ. Let u ( t ) and v ( t ) be two solutions, then w ( t ) = u ( t ) − v ( t ) satisfies k u ′ ( t ) k l u ′ ( t ) − k v ′ ( t ) k l v ′ ( t ) = − Z t ( k A u ( τ ) k β Au ( τ ) − k A v ( τ ) k β Av ( τ )) dτ − Z t ( g ( u ′ ( τ )) − g ( v ′ ( τ ))) dτ. (4.2) Applying Corollary 2.3 with f ( s ) = s l and p = l , we get kk u ′ ( t ) k l u ′ ( t ) − k v ′ ( t ) k l v ′ ( t ) k ≥ ηt ll +1 k u ′ ( t ) − v ′ ( t ) k , and applying Lemma 2.6, (4.1) and from (4.2), we now deduce k w ′ ( t ) k ≤ Ct ll +1 Z t Z τ k w ′ ( s ) k ds dτ + Ct ll +1 Z t k w ′ ( τ ) k dτ ≤ C ( T ) t − ll +1 Z t k w ′ ( τ ) k dτ. (4.3) Setting φ ( t ) = Z t k w ′ ( τ ) k dτ , by solving (4.3) on [ δ, t ] we obtain φ ( t ) ≤ φ ( δ ) e C ( T ) R tδ s − l/ ( l +1) ds , and by letting δ → w ( t ) = 0 . A similar argument gives the uniquenessfor t negative with | t | small enough. 10 roposition 4.5. For any interval J and any t ∈ J if a solution u of (1.1) satisfies u ∈ C ( J, H ) , k u ′ k l u ′ ∈ C ( J, H ) and u ( t ) = u ′ ( t ) = 0 , then u ≡ .Proof. From theorem 2.3 we know that ddt E ( t ) = − ( g ( u ′ ( t )) , u ′ ( t )) . Using (4.1), we have | ddt E ( t ) | = | ( g ( u ′ ( t )) , u ′ ( t )) | ≤ k k u ′ k l +2 ≤ k E ( t ) . Now, let t ∈ J such that E ( t ) = 0. By integration we get | E ( t ) | ≤ | E ( t ) | e k | t − t | = 0 . In this section, we suppose that ∃ η > , ∀ v, k g ( v ) k ≤ η k v k α +1 , (5.1)and ∃ η > , ∀ v, ( g ( v ) , v ) ≥ η k v k α +2 , (5.2)for some α > Theorem 5.1.
Assuming α > l , there exists a positive constant η such that if u is anysolution of (1.1) with E (0) = 0 lim inf t → + ∞ t l +2 α − l E ( t ) ≥ η. (5.3) (i) If α ≥ β (1+ l )+ lβ +2 , then there is a constant C ( E (0)) depending on E (0) such that ∀ t ≥ , E ( t ) ≤ C ( E (0)) t − l +2 α − l . (ii) If α < β (1+ l )+ lβ +2 , then there is a constant C ( E (0)) depending on E (0) such that ∀ t ≥ , E ( t ) ≤ C ( E (0)) t − ( α +1)( β +2) β − α . roof. From the definition of E ( t ) we have k u ′ ( t ) k α +2 ≤ C ( l, α ) E ( t ) α +2 l +2 , where C ( l, α ) is a positive constant, hence from (3.9) and (5.1) we deduce ddt E ( t ) ≥ − C ( l, α, η ) E ( t ) α +2 l +2 . Assuming α > l we derive ddt E ( t ) − α − ll +2 = − α − ll + 2 E ′ ( t ) E ( t ) − α +2 l +2 ≤ α − ll + 2 C ( l, α, η ) = C . By integrating, we get E ( t ) ≥ ( E (0) − α − ll +2 + C t ) − l +2 α − l , implying lim inf t → + ∞ t l +2 α − l E ( t ) ≥ η = C − l +2 α − l . Hence (5.3) is proved. Now, we show ( i ) and ( ii ), we consider the perturbed energyfunction E ε ( t ) = E ( t ) + ε ( k u k γ u, k u ′ k l u ′ ) , (5.4)where l > γ > ε > . By Young’s inequality, with the conjugate exponents l + 2 and l + 2 l + 1 , we get | ( k u k γ u, k u ′ k l u ′ ) | ≤ C k A u k (2 γ +1)( l +2) + c k u ′ k l +2 . We choose γ so that (2 γ + 1)( l + 2) ≥ β + 2, which reduces to γ ≥ β − l l + 2) . (5.5)Then, for some C > , M >
0, we have | ( k u k γ u, k u ′ k l u ′ ) | ≤ C k A u k β +2 + c k u ′ k l +2 ≤ M E ( t ) . (5.6) By using (5.6), we obtain from (5.4)(1 − M ε ) E ( t ) ≤ E ε ( t ) ≤ (1 + M ε ) E ( t ) . Taking ε ≤ M , we deduce ∀ t ≥ , E ( t ) ≤ E ε ( t ) ≤ E ( t ) . (5.7)On the other hand, we have E ′ ε ( t ) = − ( g ( u ′ ) , u ′ ) + ε (( k u k γ u ) ′ , k u ′ k l u ′ ) + ε ( k u k γ u, ( k u ′ k l u ′ ) ′ ) .
12e observe that ( k u k γ u ) ′ = (( k u k ) γ ) ′ u + k u k γ u ′ = 2 γ k u k γ − ( u, u ′ ) u + k u k γ u ′ , and we have(( k u k γ u ) ′ , k u ′ k l u ′ ) = 2 γ k u k γ − k u ′ k l ( u, u )( u ′ , u ′ ) + ( k u k γ u ′ , k u ′ k l u ′ )= 2 γ k u k γ k u ′ k l +2 + ( k u k γ u ′ , k u ′ k l u ′ ) . Then, we can deduce that E ′ ε ( t ) = − ( g ( u ′ ) , u ′ ) + ε γ k u k γ k u ′ k l +2 + ε ( k u k γ u ′ , k u ′ k l u ′ ) − ε ( k u k γ u, k A u k β Au ) − ε ( k u k γ u, g ( u ′ )) . (5.8) We now estimate the right side of (5.8),The fourth term: − ε ( k u k γ u, k A u k β Au ) = − ε k u k γ k A u k β ( u, Au )= − ε k u k γ k A u k β +2 ≤ − cε k A u k γ + β +2 . (5.9) The second term: k u k γ k u ′ k l +2 ≤ C k A u k γ k u ′ k l +2 . The third term: | ( k u k γ u ′ , k u ′ k l u ′ ) | = k u k γ k u ′ k l ( u ′ , u ′ ) ≤ C k A u k γ k u ′ k l +2 . Applying Young’s inequality, with the conjugate exponents α +2 α − l and α +2 l +2 , we have k A u k γ k u ′ k l +2 ≤ δ k A u k γ α +2 α − l + C ( δ ) k u ′ k ( l +2) α +2 l +2 . We assume ( α + 2)2 γα − l ≥ γ + β + 2 , which reduces to the condition γ ≥ ( β + 2)( α − l )2( l + 2) , (5.10)and taking δ small enough, we have for some P >
0, therefore the second and the thirdterms becomes ε γ k u k γ k u ′ k l +2 + ε ( k u k γ u ′ , k u ′ k l u ′ ) ≤ εC (2 γ + 1) k A u k γ k u ′ k l +2 ≤ ε k A u k γ + β +2 + εP k u ′ k α +2 . (5.11)Using (5.2), (5.8), (5.9) and (5.11), we have E ′ ε ( t ) ≤ ( − η + εP ) k u ′ k α +2 − ε k A u k γ + β +2 + ε k A u k γ + β +2 − ε ( k u k γ u, g ( u ′ )) . (5.12) α + 2 and α +2 α +1 and using(5.1), we have − ( k u k γ u, g ( u ′ )) ≤ δ ( k u k (2 γ +1)( α +2) + C ′ ( δ ) k g ( u ′ ) k α +2 α +1 ≤ C δ k A u k (2 γ +1)( α +2) + C ′ ( δ ) k u ′ k α +2 . This term will be dominated by the negative terms assuming(2 γ + 1)( α + 2) ≥ γ + β + 2 ⇔ ( α + 1)(2 γ + 1) ≥ β + 1 . This is equivalent to the condition γ ≥ β − α α + 1) , (5.13)and taking δ small enough, we have − ε ( k u k γ u, g ( u ′ )) ≤ ε k A u k γ + β +2 + P ′ ε k u ′ k α +2 . By replacing in (5.12), we have E ′ ε ( t ) ≤ ( − η + Qε ) k u ′ k α +2 − ε k A u k γ + β +2 , where Q = P + P ′ . By choosing ε small, we get E ′ ε ( t ) ≤ − ε (cid:16) k u ′ k α +2 + k A u k γ + β +2 (cid:17) ≤ − ε (cid:16) ( k u ′ k l +2 ) α +2 l +2 + ( k A u k β +2 ) γ + β +2 β +2 (cid:17) . (5.14) This inequality will be satisfied under the assumptions (5.5), (5.10) and (5.13) whichlead to the sufficient condition γ ≥ γ = max n β − l l + 2) , ( β + 2)( α − l )2( l + 2) , β − α α + 1) o . (5.15)We now distinguish 2 cases.(i) If α ≥ β (1 + l ) + lβ + 2 , then clearly ( β + 2)( α − l )2( l + 2) ≥ β − l l + 2) . Moreover β − α α + 1) = 12 (cid:18) β + 1 α + 1 − (cid:19) ≤ β + 1 β (1+ l )+ lβ +2 + 1 − = β − l l + 2) . In this case γ = ( β + 2)( α − l )2( l + 2) and choosing γ = γ , we find2 γ + β + 2 = α + 2 l + 2 ( β + 2) , since 2 γ + β + 2 β + 2 = 1 + α − ll + 2 , replacing in (5.14), we obtain for some ρ > E ′ ε ( t ) ≤ − ρE ( t ) α − ll +2 ≤ − ρ ′ E ε ( t ) α − ll +2 , where ρ and ρ ′ are positive constants.14ii) If α < β ( l + 1) + lβ + 2 , then ( β + 2)( α − l ) l + 2 < β − ll + 2 . Moreover β − α α + 1) − β − l l + 2) = ( β − α )( l + 2) − ( β − l )( α + 1)2( α + 1)( l + 2)= β ( l + 1) + l − α ( β + 2)2( α + 1)( l + 2) > . In this case γ = β − α α + 1) and choosing γ = γ , we find2 γ + β + 2 = ( β + 2) (cid:18) γβ + 2 (cid:19) = ( β + 2) (cid:18) β − α ( α + 1)( β + 2) (cid:19) , (5.16)since γ > ( β + 2)( α − l ) l + 2 , we have2 γ + β + 2 β + 2 = 1 + 2 γβ + 2 > α − ll + 2 = α + 2 l + 2 , replacing in (5.14), we obtain E ′ ε ( t ) ≤ − δε ( k u ′ k l +2 + k A u k β +2 ) γ + β +2 β +2 , for some δ >
0. Using (5.16), we have E ′ ε ( t ) ≤ − ρE (1+ β − α ( α +1)( β +2) ) ≤ − ρ ′ E ε ( t ) (1+ β − α ( α +1)( β +2) ) , where ρ and ρ ′ are positive constants. In the case where α < β (1+ l )+ lβ +2 , Theorem 5.1 gives two different decay rates for thelower and the upper estimates of the energy. In the scalar case, this fact was explainedin [1] by the existence of two (and only two) different decay rates of the solutions, cor-responding precisely to the lower and upper estimates. The solutions behaving as thelower estimate were called “fast solutions” and those behaving as the upper estimatewere called “slow solutions.” Moreover in the scalar case, it was shown that the set ofinitial data giving rise to “slow solutions” has non-empty interior in the phase space IR . In the general case, by reducing the problem to a related scalar equation, it is ratherimmediate to show the coexistence of slow and fast solutions in the special case of powernonlinearities. More precisely we have 15 roposition 6.1.
Let α < β (1+ l )+ lβ +2 and c > . Then the equation ( k u ′ k l u ′ ) ′ + k A u k β Au + c k u ′ k α u ′ = 0 has an infinity of “fast solutions” with energy comparable to t − l +2 α − l and an infinity of“slow solutions” with energy comparable to t − ( α +1)( β +2) β − α as t tends to infinity.Proof. Let λ > A and Aϕ = λϕ with k ϕ k = 1 . Let ( v , v ) ∈ IR and v be the solution of ( | v ′ | l v ′ ) ′ + C | v | β v + C | v ′ | α v ′ = 0 , where C , C are positive constants to be chosen later. Then u ( t ) = v ( t ) ϕ satisfies( k u ′ k l u ′ ) ′ + k A u k β Au + c k u ′ k α u ′ = ( | v ′ | l v ′ ) ′ k ϕ k l ϕ + | v | β v k A ϕ k β Aϕ + c | v ′ | α v ′ k ϕ k α ϕ = [( | v ′ | l v ′ ) ′ + λ β +1 | v | β v + c | v ′ | α v ′ ] ϕ. Choosing C = λ β +1 and C = c , u ( t ) = v ( t ) ϕ becomes a solution of the vector equa-tion. The existence of an infinity of “fast solutions” and an infinity of “slow solutions”are then an immediate consequence of the same result for the scalar equation proven in [1] . Remark 6.2.
In the special case A = λI , we can take for ϕ any vector of the sphere k ϕ k = 1 . We obtain in this way an open set of slow solutions in R N +1 correspondingto the initial data of the form ( v ϕ, v ϕ ) . Actually, in the general case, by generalizinga modified energy method introduced in [3] , under the additional condition l <
Theorem 6.3.
Assume that g satisfies (5.2) and l < , l < α < β (1+ l )+ lβ +2 . Then, thereexist a nonempty open set S ⊂ H × H and a constant M such that, for every ( u , u ) ∈ S ,the unique global solution of equation (1.1) with initial data ( u , u ) satisfies k u ( t ) k ≥ M (1 + t ) α +1 β − α ∀ t ≥ . (6.1) Proof.
Assuming ( u , u ) ∈ H × H and u = 0, we consider the following constants σ := (cid:18) ( β + 2)( l + 1) l + 2 k u k l +2 + k A u k β +2 (cid:19) β +2 , and σ := k u k k A u k ( β − αα +1 +1 ) . For any ε > ε >
0, the set
S ⊂ H × H of initial data such that σ < ε , σ < ε isclearly a nonempty open set which contains at least all pairs ( u , u ) with u = 0 and16 = 0 with k u k small enough. We claim that if ε and ε are small enough, for any( u , u ) ∈ S , the global solution of (1.1) satisfies (6.1).First of all, from (3.9) and (5.2), we see that ddt E ( t ) ≤ − η (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) α +2 < , hence E ( t ) ≤ E (0) for every t ≥
0, and from (1.2), we deduce that k A u ( t ) k ≤ [( β + 2) E (0)] β +2 = σ ∀ t ≥ . (6.2)Then, we shall establish that if ε and ε are small enough, we have u ( t ) = 0 ∀ t ≥ , (6.3)and for some C > k u ′ ( t ) k k A u ( t ) k ( β − αα +1 +1 ) ≤ C ∀ t ≥ . (6.4)Assuming this inequality, it follows immediately that u is a slow solution. Indeed let usset y ( t ) := k u ( t ) k . We observe that (cid:12)(cid:12) y ′ ( t ) (cid:12)(cid:12) = 2 (cid:12)(cid:12)(cid:0) u ′ ( t ) , u ( t ) (cid:1)(cid:12)(cid:12) ≤ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) · k u ( t ) k≤ k u ′ ( t ) kk A u ( t ) k β − αα +1 +1 · k A u ( t ) k β − αα +1 +1 k u ( t ) k≤ √ C k u ( t ) k β − αα +1 +2 ≤ √ C | y ( t ) | β − α α +1) +1 , and in particular y ′ ( t ) ≥ − √ C | y ( t ) | β − α α +1) +1 ∀ t ≥ . Taking into account that (cid:18) y ( t ) − β − α α +1) (cid:19) ′ = − β − α α + 1) y ( t ) − β − α α +1) − y ′ ( t ) , we have (cid:18) y ( t ) − β − α α +1) (cid:19) ′ ≤ √ C β − αα + 1 . Integrating between 0 and t and since y (0) >
0, we deduce that there exists a constant M such that y ( t ) − β − α α +1) ≤ M ( t + 1) ∀ t ≥ . This inequality concludes the proof. So we are left to prove (6.3) and (6.4).17et us set T := sup { t ≥ ∀ τ ∈ [0 , t ] , u ( τ ) = 0 } . Since u (0) = 0, we have that T >
T < + ∞ , then u ( T ) = 0.Let us consider, for all t ∈ (0 , T ), the energy H ( t ) := k u ′ ( t ) k k A u ( t ) k γ , (6.5)where γ := β − αα + 1 + 1.If u ′ ( t ) = 0, then we claim that H is differentiable at t with H ′ ( t ) = 0. Indeed bythe equation (cid:16) k u ′ k l u ′ (cid:17) ′ + k A u k β Au + g ( u ′ ) = 0 , we find that k u ′ k l u ′ ( s ) = O ( s − t ) for s close to t , then since l < k u ′ ( t ) k = O ( s − t ), thereby proving the claim.If u ′ ( t ) = 0 , then again H is differentiable at t with H ′ ( t ) = ddt (cid:16)(cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l − ddt (cid:16) k A u ( t ) k γ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l H ( t ) . Taking into account (1.2) and (3.9), we deduce H ′ ( t ) = − ρ ( g ( u ′ ( t )) , u ′ ( t )) k A u ( t ) k γ k u ′ ( t ) k l − ρ k A u ( t ) k β − γ ( u ′ ( t ) , Au ( t )) k u ′ ( t ) k l − ddt (cid:16) k A u ( t ) k γ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l H ( t ) =: H + H + H , (6.6)where ρ := l + 2 l + 1 .Let us estimate H , H and H . Using (5.2) and (6.5), we observe that H ≤ − ρ η k u ′ ( t ) k α +2 − l k A u ( t ) k γ = − ρ η k u ′ ( t ) k k A u ( t ) k γ ! α +2 − l k A u ( t ) k γ ( α +2 − l ) k A u ( t ) k γ = − ρ η H α +2 − l ( t ) k A u ( t ) k γ ( α − l ) . (6.7)In order to estimate H , we use Young’s inequality applied with the conjugate exponents α +2 − l − l and α +2 − lα +1 , | H | ≤ ρ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) − l · k A u ( t ) k β +1 k A u ( t ) k γ = ρ k u ′ ( t ) k − l k A u ( t ) k γ (1 − l ) α +2 − l · k A u ( t ) k γ (1 − l ) α +2 − l + β +1 k A u ( t ) k γ ≤ δ k u ′ ( t ) k − l k A u ( t ) k γ (1 − l ) α +2 − l ! α +2 − l − l + δ (cid:18) k A u ( t ) k γ (1 − l ) α +2 − l + β +1 − γ (cid:19) α +2 − lα +1 , δ := ρ η (cid:18) − lα + 2 − l (cid:19) and δ := ρη (cid:18) α + 1 α + 2 − l (cid:19) , and taking into account that γ = β +1 α +1 , we deduce | H | ≤ δ k u ′ ( t ) k k A u ( t ) k γ ! α +2 − l k A u ( t ) k γ ( α +2 − l ) k A u ( t ) k γ + δ k A u ( t ) k γ ( α − l ) = δ H α +2 − l ( t ) k A u ( t ) k γ ( α − l ) + δ k A u ( t ) k γ ( α − l ) . (6.8)For simplicity in the sequel we shall write l/ m > . In order to estimate H , we use (6.5) and we have that H = − ddt (cid:16) k A u ( t ) k γ (1+ m ) H ( t ) m (cid:17) k A u ( t ) k γ k u ′ ( t ) k l H ( t ) = − mH ( t ) m H ′ ( t ) k A u ( t ) k γm k u ′ ( t ) k l − γ (1 + m ) H ( t ) m +1 k A u ( t ) k γm − k u ′ ( t ) k l (cid:0) u ′ ( t ) , Au ( t ) (cid:1) =: H + H . We observe that taking into account (6.5), we have H = − mH ′ ( t ) , (6.9)and | H | ≤ γ (1 + m ) H ( t ) m +1 k A u ( t ) k γm k u ′ ( t ) k l · k u ′ ( t ) kk A u ( t ) k = 2 γ (1 + m ) H ( t ) k A u ( t ) k γ − . We observe that the hypothesis α < β (1 + l ) + lβ + 2 , implies γ − > γ ( α − l ) , and taking into account (6.2), we obtain | H | ≤ γ (1 + m ) σ γ − − γ ( α − l )0 H ( t ) k A u ( t ) k γ ( α − l ) . (6.10)Then, taking into account (6.7)-(6.10) in (6.6), we deduce that H ′ ( t ) ≤ k A u ( t ) k γ ( α − l ) (cid:20) − b ρ η H α +2 − l ( t ) + b ρη + 2 γσ γ − − γ ( α − l )0 H ( t ) (cid:21) , (6.11)where b ρ := ρ ( α +1)(1+ m )( α +2 − l ) . 19ere we observe that (6.11) is still valid if u ′ ( t ) = 0, since then H ( t ) = 0 = H ′ ( t )and the RHS is non-negative. Now, let h ( s, Γ) = − b ρ η s α +2 − l + b ρη + 2 γ Γ s , where Γ := σ γ − − γ ( α − l )0 . We observe that h ((2 /η ) / ( α +2 − l ) , Γ) = − b ρη + 2 γ Γ (cid:0) /η (cid:1) α +2 − l . In particular h (cid:16) (2 /η ) / ( α +2 − l ) , Γ (cid:17) → − b ρη < , if Γ → . Let us therefore assume that σ is sufficiently small to achieve h ((2 /η ) / ( α +2 − l ) , σ γ − − γ ( α − l )0 ) < . We claim that if H (0) = σ < ε := (2 /η ) / ( α +2 − l ) , then H ( t ) is bounded for all t ∈ (0 , T ). Indeed if H is not bounded for all t ∈ (0 , T ), then there exists ¯ T ∈ (0 , T )such that H ( ¯ T ) = ε and H ( t ) < ε for all t ∈ (0 , ¯ T ). By (6.11) and taking into accountthat u ∈ C ( R + , H ) and k u ′ k l u ′ ∈ C ( R + , H ), we have H ′ ( ¯ T ) ≤ k A u ( ¯ T ) k γ ( α − l ) h ( ε , σ γ − − γ ( α − l )0 ) ≤ (cid:13)(cid:13) u ( ¯ T ) (cid:13)(cid:13) γ ( α − l ) h ( ε , σ γ − − γ ( α − l )0 ) < , then since ¯ T ∈ (0 , T ) implies (cid:13)(cid:13) u ( ¯ T ) (cid:13)(cid:13) > H decreases near ¯ T . This contradicts thedefinition of ¯ T and we can claim that H is bounded for all t ∈ (0 , T ).Finally, we claim that T = + ∞ . Let us assume by contradiction that this is not thecase. Then, taking into account that H is bounded for all t ∈ [0 , T ), we observe that (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) ≤ C k A u ( t ) k γ ≤ C k u ( t ) k γ ∀ t ∈ [0 , T ) . Therefore, from the continuity of the vector ( u ′ , u ) with values in H × H it nowfollows that u ( T ) = 0 implies u ′ ( T ) = 0, hence u ≡ T = + ∞ , we obtain (6.3) and (6.4), and the conclusion follows.Our next result is the generalization to the vector case of the slow-fast alternativeestablished in [1] for the scalar equation. By relying on a technique introduced by Ghisi,Gobbino and Haraux [4], we prove that all non-zero solutions to (1.1) are either slowsolutions or fast solutions. Theorem 6.4 (The slow-fast alternative) . Assume that g is locally Lipschitz continuousand satisfies (5.2) and l < , l < α < β (1+ l )+ lβ +2 . Let u ( t ) be a global solution to (1.1).Then, one and only one of the following statements apply: . (Fast solutions) There exist some positive constants η , C and T such that k u ′ ( t ) k ≥ η k A u ( t ) k ( β +2 l +2 ) , ∀ t ≥ T, and E ( t ) ≤ C ( t + 1) l +2 α − l , ∀ t ≥ . (6.12)
2. (Slow solutions) There exist some positive constants M , c and T such that k u ′ ( t ) k ≤ M k A u ( t ) k ( β +2 l +2 ) , ∀ t ≥ T, and k u ( t ) k ≥ c (1 + t ) α +1 β − α , ∀ t ≥ . (6.13) Proof.
Fast solutions:
We first establish that if for some η > k u ′ ( t ) k ≥ η k A u ( t ) k ( β +2 l +2 ) , ∀ t ≥ T, (6.14)then (6.12) holds.Using (1.2), we obtain l + 1 l + 2 k u ′ ( t ) k l +2 = − β + 2 k A u ( t ) k β +2 + E ( t ) ≥ E ( t ) − δ k u ′ ( t ) k l +2 , where δ := h ( β + 2) η l +22 i − . Then k u ′ ( t ) k α +2 ≥ (cid:18) l + 1 l + 2 + δ (cid:19) − α +2 l +2 E ( t ) α +2 l +2 . From the energy identity, we deduce E ′ ( t ) ≤ − b CE ( t ) α − ll +2 +1 , where b C := η (cid:18) l + 1 l + 2 + δ (cid:19) − α +2 l +2 . Integrating between T and t and since E ( t ) ≥
0, wededuce that there exists a constant C such that E ( t ) ≤ C ( t − T ) − l +2 α − l , ∀ t ≥ T + 1 . The result follows immediately.
Slow solutions:
In addition to the already defined number γ = β − αα + 1 + 1 = β + 1 α + 1 ,let us introduce ˆ γ = β − ll + 2 + 1 = β + 2 l + 2 + 1. The condition α < β (1 + l ) + lβ + 2 , β − ll + 2 < β − αα + 1 . Therefore ˆ γ < γ.
In this section, we consider the energies H ( t ) and K ( t ) := k u ′ ( t ) k k A u ( t ) k γ , (6.15)defined whenever u ( t ) = 0. In this part of the proof, it remains to consider the casewhere (6.14) is false for every η . Then, assuming the solution to be non-trivial thereexists a sequence t n → + ∞ such that u ( t n ) = 0 and K ( t n ) −→
0. Therefore, forany ε > N ( ε ) ∈ N such that u ( t n ) = 0 and K ( t n ) ≤ ε , ∀ n ≥ N ( ε ) . (6.16)Under this assumption, we prove that a similar estimate holds true for all sufficientlylarge times, namely for some n ∈ N u ( t ) = 0 and K ( t ) ≤ M, ∀ t ≥ t n , (6.17)for a suitable constant M ≥ ε . In order to achieve this property, we select anotherpositive number, ε , to be fixed later in the proof, and we consider n ∈ N large enoughto achieve the additional condition k A u ( t ) k ≤ ε . Let us set ¯ T := sup { t ≥ t n : ∀ τ ∈ [ t n , t ] , u ( τ ) = 0 } . By (6.16), we deduce that u ( t n ) = 0 and we have that ¯ T >
0. If ¯
T < + ∞ , then u ( ¯ T ) = 0. Let us consider the energy K , which is defined by (6.15) for every t ∈ [ t n , ¯ T ).If u ′ ( t ) = 0, then we claim that K is differentiable at t with K ′ ( t ) = 0. Indeed bythe equation we find that k u ′ k l u ′ ( s ) = O ( s − t ) for s close to t , then since l < k u ′ ( t ) k = O ( s − t ), thereby proving the claim.If u ′ ( t ) = 0, then again K is differentiable at t with: K ′ ( t ) = ddt (cid:16)(cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l − ddt (cid:16) k A u ( t ) k γ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l K ( t ) . K ′ ( t ) ≤ − ρ η k u ′ ( t ) k α +2 − l k A u ( t ) k γ − ρ k A u ( t ) k β − γ ( u ′ ( t ) , Au ( t )) k u ′ ( t ) k l − ddt (cid:16) k A u ( t ) k γ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l K ( t ) =: K + K + K , (6.18)where ρ := l + 2 l + 1 .Let us estimate K , K and K . Using (6.15), we observe K = − ρ η k u ′ ( t ) k k A u ( t ) k γ ! α +2 − l k A u ( t ) k ˆ γ ( α +2 − l ) k A u ( t ) k γ = − ρ η K α +2 − l ( t ) k A u ( t ) k ˆ γ ( α − l ) . (6.19)In order to estimate K , we use Young’s inequality applied with the conjugate exponents α +2 − l − l and α +2 − lα +1 , (cid:12)(cid:12) K (cid:12)(cid:12) ≤ ρ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) − l · k A u ( t ) k β +1 k A u ( t ) k γ ≤ k A u ( t ) k γ (cid:18) δ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) α +2 − l + δ k A u ( t ) k ( β +1)( α +2 − l ) α +1 (cid:19) , where δ := ρ η (cid:18) − lα + 2 − l (cid:19) and δ := ρη (cid:18) α + 1 α + 2 − l (cid:19) , and taking into account that γ = β +1 α +1 , we deduce (cid:12)(cid:12) K (cid:12)(cid:12) ≤ δ k u ′ ( t ) k k A u ( t ) k γ ! α +2 − l k A u ( t ) k ˆ γ ( α +2 − l ) k A u ( t ) k γ + δ k A u ( t ) k ( β +1) ( α +2 − lα +1 − l +2 ) − l +2 = δ K α +2 − l ( t ) k A u ( t ) k ˆ γ ( α − l ) + δ k A u ( t ) k µ , where µ := ( β + 1) (cid:16) α +2 − lα +1 − l +2 (cid:17) − l +2 = ( β +1)( α +2 − l ) α +1 − γ .By simple calculations, we deduce from α < β (1+ l )+ lβ +2 thatˆ γ ( α − l ) < µ, therefore (cid:12)(cid:12) K (cid:12)(cid:12) ≤ k A u ( t ) k ˆ γ ( α − l ) ( δ K α +2 − l ( t ) + δ k A u ( t ) k ν ) , (6.20)with ν = µ − ˆ γ ( α − l ) >
0. 23or simplicity in the sequel we shall write l/ m > . (6.21)In order to estimate K , we use (6.15) and we have that K = − ddt (cid:16) k A u ( t ) k γ (1+ m ) K m ( t ) (cid:17) k A u ( t ) k γ k u ′ ( t ) k l K ( t ) = − mK m ( t ) K ′ ( t ) k A u ( t ) k γm k u ′ ( t ) k l − γ (1 + m ) K m +1 ( t ) k A u ( t ) k γm − k u ′ ( t ) k l (cid:0) u ′ ( t ) , Au ( t ) (cid:1) =: K + K . We observe that taking account of (6.15), we have K = − mK ′ ( t ) , (6.22)and (cid:12)(cid:12) K (cid:12)(cid:12) ≤ γ (1 + m ) K m +1 ( t ) k A u ( t ) k γm k u ′ ( t ) k l · k u ′ ( t ) kk A u ( t ) k = 2ˆ γ (1 + m ) K ( t ) k A u ( t ) k ˆ γ − . We observe that our condition on α can be written in the form α ( β + 2) < ( β + 2)(1 + l ) − ( l + 2) ⇒ ( β + 2)(1 + l − α ) > l + 2 ⇒ ˆ γ (1 + l − α ) > ⇒ ˆ γ − > ˆ γ ( α − l ) . Using (6.2) we therefore obtain (cid:12)(cid:12) K (cid:12)(cid:12) ≤ C K ( t ) k A u ( t ) k ˆ γ ( α − l ) . (6.23)Then, taking into account (6.19)-(6.23) in (6.18), we deduce that K ′ ( t ) ≤ k A u ( t ) k ˆ γ ( α − l ) h − b ρ η K α +2 − l ( t )+ C k A u ( t ) k ν + C K ( t ) i , (6.24)where b ρ := ρ ( α +1)(1+ m )( α +2 − l ) .Here we observe that (6.24) is still valid if u ′ ( t ) = 0, since then K ( t ) = 0 = K ′ ( t )and the RHS is non-negative. Now, let h ( s ) = − b ρ η s α +2 − l + C ε ν + C s , and ˜ h ( s ) = − b ρ η s α +2 − l + C s . Since α < β (1+ l )+ lβ +2 , we have α < l ( β + 1) β + 2 = ⇒ α + 2 − l < .
24e observe that ˜ h ( s ) < , if s → . Fixing ε sufficiently small to achieve ˜ h ( ε ) < ε so that h ( ε ) < . Now, if K is not bounded for all t ∈ ( t n , ¯ T ), then there exists T ∈ ( t n , ¯ T ) suchthat K ( T ) = ε and K ( t ) < ε for all t ∈ ( t n , T ).By (6.24) and taking into account that u ∈ C ( R + , H ) and k u ′ k l u ′ ∈ C ( R + , H ), wehave K ′ ( T ) ≤ k A u ( T ) k ˆ γ ( α − l ) h ( ε ) ≤ k u ( T ) k ˆ γ ( α − l ) h ( ε ) < , then since T ∈ ( t n , ¯ T ) implies k u ( T ) k > K decreases near T . This contradicts thedefinition of T and we can claim that K is bounded for all t ∈ ( t n , ¯ T ).We claim that ¯ T = + ∞ . Let us assume by contradiction that this is not the case.Then, taking into account that K is bounded for all t ∈ [ t n , ¯ T ), we observe that (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) ≤ M k A u ( t ) k γ , ∀ t ∈ [ t n , ¯ T ) . Therefore, from the continuity of the vector ( u ′ , u ) with values in H × H it nowfollows that u ( ¯ T ) = 0 implies u ′ ( ¯ T ) = 0, hence u ≡ T = + ∞ . Then, we obtain (6.17).So we are left to prove that for some M > k u ′ ( t ) k k A u ( t ) k ( β − αα +1 +1 ) ≤ M ∀ t ≥ t n . (6.25)Let us consider the energy H , which is defined by (6.5) for every t ≥ t n .Assuming first u ′ ( t ) = 0, we can differentiate H at t, and we find H ′ ( t ) = ddt (cid:16)(cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l − ddt (cid:16) k A u ( t ) k γ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l H ( t ) . Taking into account (1.2) and the energy identity, we deduce H ′ ( t ) ≤ − ρ η k u ′ ( t ) k α +2 − l k A u ( t ) k γ − ρ k A u ( t ) k β − γ ( u ′ ( t ) , Au ( t )) k u ′ ( t ) k l − ddt (cid:16) k A u ( t ) k γ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) l (cid:17) k A u ( t ) k γ k u ′ ( t ) k l H ( t ) =: H + H + H , (6.26)where ρ := l + 2 l + 1 .Let us estimate H , H and H . Using (6.5), we observe that H = − ρ η k u ′ ( t ) k k A u ( t ) k γ ! α +2 − l k A u ( t ) k γ ( α +2 − l ) k A u ( t ) k γ = − ρ η H α +2 − l ( t ) k A u ( t ) k γ ( α − l ) . (6.27)25n order to estimate H , we use Young’s inequality applied with the conjugate exponents α +2 − l − l and α +2 − lα +1 , (cid:12)(cid:12) H (cid:12)(cid:12) ≤ ρ (cid:13)(cid:13) u ′ ( t ) (cid:13)(cid:13) − l · k A u ( t ) k β +1 k A u ( t ) k γ = ρ k u ′ ( t ) k − l k A u ( t ) k γ (1 − l ) α +2 − l · k A u ( t ) k γ (1 − l ) α +2 − l + β +1 k A u ( t ) k γ ≤ δ k u ′ ( t ) k − l k A u ( t ) k γ (1 − l ) α +2 − l ! α +2 − l − l + δ (cid:18) k A u ( t ) k γ (1 − l ) α +2 − l + β +1 − γ (cid:19) α +2 − lα +1 , where δ := ρ η (cid:18) − lα + 2 − l (cid:19) and δ := ρη (cid:18) α + 1 α + 2 − l (cid:19) , and taking into account that γ = β +1 α +1 , we deduce (cid:12)(cid:12) H (cid:12)(cid:12) ≤ δ k u ′ ( t ) k k A u ( t ) k γ ! α +2 − l k A u ( t ) k γ ( α +2 − l ) k A u ( t ) k γ + δ k A u ( t ) k γ ( α − l ) = δ H α +2 − l ( t ) k A u ( t ) k γ ( α − l ) + δ k A u ( t ) k γ ( α − l ) . (6.28)In order to estimate H , using (6.21) and (6.5), we have that H = − ddt (cid:16) k A u ( t ) k γ (1+ m ) H m ( t ) (cid:17) k A u ( t ) k γ k u ′ ( t ) k l H ( t ) = − mH m ( t ) H ′ ( t ) k A u ( t ) k γm k u ′ ( t ) k l − γ (1 + m ) H m +1 ( t ) k A u ( t ) k γm − k u ′ ( t ) k l (cid:0) u ′ ( t ) , Au ( t ) (cid:1) =: H + H . We observe that taking into account (6.5), we have H = − mH ′ ( t ) , (6.29)and using (6.15), we obtain (cid:12)(cid:12) H (cid:12)(cid:12) ≤ γ (1 + m ) H m +1 ( t ) k A u ( t ) k γm k u ′ ( t ) k l · k u ′ ( t ) kk A u ( t ) k (6.30)= 2 γ (1 + m ) H ( t ) K ( t ) k A u ( t ) k ˆ γ − . We observe, after some calculations, thatˆ γ − − γ ( α − l ) > ⇐⇒ α ( β + 2)(1 + l ) < l ( β + 1)( l + 2) + β − l, which reduces easily to the condition α < β (1 + l ) + lβ + 2 . Then, taking into account (6.27)-(6.30) in (6.26), we deduce that H ′ ( t ) ≤ k A u ( t ) k γ ( α − l ) (cid:20) − b ρ η H α +2 − l ( t ) + b ρη + 2 γH ( t ) K ( t ) k A u ( t ) k ˆ γ − − γ ( α − l ) (cid:21) , b ρ := ρ ( α +1)(1+ m )( α +2 − l ) . Now, we estimate the third term. We have that K is boundedfor all t ≥ t n . Then, using Young’s inequality applied with the conjugate exponents α +2 − l and α +2 − lα − l ,2 γH ( t ) K ( t ) k A u ( t ) k ˆ γ − − γ ( α − l ) ≤ f M H ( t ) ≤ b ρ η H α +2 − l ( t ) + 4( α − l ) b ρ η ( α + 2 − l ) f M α +2 − lα − l , where f M := 2 γM ε ˆ γ − − γ ( α − l )2 .Then, we obtain H ′ ( t ) ≤ k A u ( t ) k γ ( α − l ) (cid:20) − b ρ η H α +2 − l ( t ) + b K (cid:21) , where b K := b ρη + 4( α − l ) b ρ η ( α + 2 − l ) f M α +2 − lα − l . If u ′ ( t ) = 0, since as previously H is differ-entiable at t with H ′ ( t ) = 0 = H ( t ), the inequality is still valid.It remains to prove that H ( t ) is bounded for all t ≥ t n . Indeed if H is not boundedfor all t ≥ t n , then there exist T ≥ t n such that T := inf { t ≥ t n : ∀ τ ∈ [ t n , t ] , b ρ η H α +2 − l ( τ ) < b ρ η H α +2 − l ( t n ) + b K + 1 } . We have H ′ ( T ) ≤ k A u ( T ) k γ ( α − l ) (cid:20) − b ρ η H α +2 − l ( T ) + b K (cid:21) = −k A u ( T ) k γ ( α − l ) (cid:20) b ρ η H α +2 − l ( t n ) (cid:21) < , and since T ≥ t n , implies u ( T ) = 0, H decreases near T . This contradicts thedefinition of T and we can claim that H is bounded for all t ≥ t n . The end of theproof is identical to that of Theorem 6.3. Remark 6.5.
In the special case A = λI , by using the scalar equation we can provethe existence of an N - dimensional variety of fast solutions. Actually, by Theorem 5.4,(4) of [4] we know that the same result is true in general for the non-singular linearlydamped equation u ′′ + u ′ + k A / u k β Au = 0 . The question now arises of whether the dimensionality of the variety of fast solutions isexactly equal to N or possibly larger. The special case of the equation u ′′ + u ′ + k u k β u = 0 (6.31)shows it is exactly equal to N at least in some cases. Indeed we claim that in this case,all fast solutions have their N components proportional to the same fast solution of arelated scalar equation. 27 roof. Given a solution u of (6.31), any component y of u satisfies y ′′ + y ′ + k u k β y = 0 . Considering two different components ( y, z ) we obtain that the Wronskian function w ( t ) := ( y ′ z − yz ′ )( t ) is a solution of w ′ + w = 0 . Hence w ( t ) = w (0) exp( − t ). Since on the other hand w , being a quadratic combinationof 4 functions decaying like exp( − t ), has to decay like exp( − t ), the only possibility is w (0) = 0, hence w = 0. Hence any two components are proportional. Let us set y j = a j y, where y is one of the non-zero components. Then we have y ′′ + y ′ + m k y k β y = 0with m := { P Nj =1 a j } β/ . Let us introduce φ := µy where µ > φ satisfies φ ′′ + φ ′ + µm k y k β y = 0 or φ ′′ + φ ′ + µ − β m k φ k β φ = 0 , so that the choice µ = m /β leads to φ ′′ + φ ′ + k φ k β φ = 0 . Finally we have y j = b j φ, with b j = m − /β a j and φ a fast solution of the scalar equation. Since the set of such fastsolutions φ is invariant by translation in t , for the moment we reduced the dimensionto N + 1 instead of N and we know that this is not relevant since in the scalar casethe dimension is 1. Actually the numbers b j are constrained by an additional condition:indeed we have for all j the equation y ′′ j + y ′ j + k u k β y j = b j ( φ ′′ + φ ′ + k u k β φ ) = 0 , and since at least one of the b j is not 0, we obtain φ ′′ + φ ′ + k u k β φ = 0 . Now k u k β = n N X j =1 b j o β/ k φ k β . Since slow solutions do not vanish for t large we conclude that P Nj =1 b j = 1, the ad-ditional constraint we were looking for, reducing the dimension to N . The set of slowsolutions has the structure of a 1D curve fibrated by the ( N − g is non-linear or A is not a multiple of the identity, we haveno information on the dimension of this set. This is one of the last open problems ofthis theory. 28 cknowledgments M. Anguiano has been supported by Junta de Andaluc´ıa (Spain), Proyecto de Excelen-cia P12-FQM-2466, and in part by European Commission, Excellent Science-EuropeanResearch Council (ERC) H2020-EU.1.1.-639227.
References [1] M. Abdelli and A. Haraux,
Global behavior of the solutions to a class of nonlinearsecond order ODE’s,
Nonlinear Analysis, (2014), 18-73.[2] F. Aloui, I. Ben Hassen and A. Haraux Compactness of trajectories to some nonlin-ear second order evolution equations and applications,
Journal de Mathematiquespures et appliquees ,
No, 3 (2013), 295-326.[3] M. Ghisi, M. Gobbino and A. Haraux,
Optimal decay estimates for the generalsolution to a class of semi-linear dissipative hyperbolic equations,
J. Eur. Math.Soc. (JEMS) (2016), no. 9, 1961–1982.[4] M. Ghisi, M. Gobbino and A. Haraux, Finding the exact decay rate of all solu-tions to some second order evolution equation with dissipation,
J. Funct. Anal. (2016), no. 9, 2359–2395.[5] A. Haraux,
Sharp decay estimates of the solutions to a class of nonlinear secondorder ODE’s,
Analysis and Applications, (2011), 49-69.[6] A. Haraux, Slow and fast decay of solutions to some second order evolution equa-tions,
J. Anal. Math.95